Niemytzki Plane is Topology

Theorem
Niemytzki plane is a topological space.

Proof
By definition $T = \left({S, \tau}\right)$ is the Niemytzki plane if

where $B_r \left({x, y}\right)$ denotes the open $r$-ball of $\left({x, y}\right)$ in the $\R^2$ Euclidean space.

According to Topology Defined by Neighborhood System it should be proved
 * $(N0): \quad \forall \left({x, y}\right) \in S: \mathcal B\left({x, y}\right)$ is non-empty set of subsets of $S$
 * $(N1): \quad \forall \left({x, y}\right) \in S, U \in \mathcal B\left({x, y}\right): \left({x, y}\right) \in U$
 * $(N2): \quad \forall \left({x, z}\right) \in S, U \in \mathcal B\left({x, z}\right), \left({y, s}\right) \in U:\exists V \in \mathcal B\left({y, s}\right): V \subseteq U$
 * $(N3): \quad \forall \left({x, y}\right) \in S, U_1, U_2 \in \mathcal B\left({x, y}\right): \exists U \in \mathcal B\left({x, y}\right): U \subseteq U_1 \cap U_2$

Ad $(N0)$:

Let $\left({x, y}\right) \in S$.

In both cases: $y > 0$ and $y = 0$, by $(2)$ and $(3)$:
 * $\mathcal B \left({x, y}\right)$ is non-empty

Let $U \in \mathcal B \left({x, y}\right)$.

In a case when $y > 0$:
 * $\exists r > 0: U = B_r\left({x, y}\right) \cap S$

By Intersection is Subset:
 * $U$ is a subset of $S$.

In a case when $y = 0$:
 * $\exists r > 0: U = B_r\left({x, r}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By definition of subset:
 * $B_r\left({x, r}\right) \subseteq S$ and $\left\{{\left({x, 0}\right)}\right\} \subseteq S$

By Union is Smallest Superset:
 * $U$ is a subset of $S$.

Ad $(N1)$:

Let $\left({x, y}\right) \in S$.

Let $U \in \mathcal B \left({x, y}\right)$.

In a case when $y > 0$ by $(2)$:
 * $\exists r > 0: U = B_r\left({x, y}\right) \cap S$

By metric spce axioms:
 * $d\left({\left({x, y}\right), \left({x, y}\right)}\right) = 0$

where $d$ denotes the distance function of $\R^2$ Euclidean space.

By definition of open ball:
 * $\left({x, y}\right) \in B_r\left({x, y}\right)$

Thus by definition of intersection:
 * $\left({x, y}\right) \in U$

In a case when $y = 0$ by $(3)$:
 * $\exists r > 0: U = B_r\left({x, r}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By definition of singleton:
 * $\left({x, 0}\right) \in \left\{{\left({x, 0}\right)}\right\}$

Thus by definition of union:
 * $\left({x, 0}\right) \in U$

Thus in both cases:
 * $\left({x, 0}\right) \in U$

Ad $(N2)$:

Let $\left({x, 0}\right) \in S, U \in \mathcal B \left({x, y}\right), \left({z, s}\right) \in U$.

In a case when $y > 0$ by $(2)$:
 * $\exists r > 0: U = B_r\left({x, y}\right) \cap S$

By definition of intersection:
 * $\left({z, s}\right) \in B_r\left({x, y}\right)$

By definition of open ball:
 * $d \left({\left({x, y}\right), \left({z, s}\right)}\right) < r$

In a subcase when $s > 0$ define
 * $r' := r-d \left({\left({x, y}\right), \left({z, s}\right)}\right)$

and
 * $V := B_{r'}\left({z, s}\right) \cap S$

By $(2)$:
 * $V \in \mathcal B \left({z, s}\right)$

By proof of Open Ball of Point Inside Open Ball:
 * $B_{r'} \left({z, s}\right) \subseteq B_r\left({x, y}\right)$

Thus by Set Intersection Preserves Subsets/Corollary:
 * $V \subseteq U$

In a subcase when $s = 0$ define
 * $r'' := \displaystyle \frac {r-d \left({\left({x, y}\right), \left({z, s}\right)}\right)} 2$

and
 * $V := B_{r}\left({z, r}\right) \cup \left\{{\left({z, 0}\right)}\right\}$

By $(3)$:
 * $V \in \mathcal B \left({z, s}\right)$

By Open Ball is Subset of Open Ball:
 * $B_{r}\left({z, r}\right) \subseteq B_r\left({x, y}\right)$

and
 * $B_{r}\left({z, r}\right) \subseteq S$

By Intersection is Largest Subset:
 * $B_{r}\left({z, r}\right) \subseteq U$

By definition of subset:
 * $\left\{{\left({z, 0}\right)}\right\} \subseteq U$

Thus by Union is Smallest Superset:
 * $V \subseteq U$

In a case when $y = 0$ by $(3)$:
 * $\exists r > 0: U = B_r\left({x, r}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By definition of onion and singleton:
 * $\left({z, s}\right) \in B_r\left({x, r}\right)$ or $\left({z, s}\right) = \left({x, 0}\right)$

In a subcase when $\left({z, s}\right) = \left({x, 0}\right)$ defint
 * $V := U$

By $(3)$:
 * $V \in \mathcal B \left({z, s}\right)$

Thus by Set is Subset of Itself:
 * $V \subseteq U$

In a subcase when $\left({z, s}\right) \in B_r\left({x, r}\right)$ define
 * $r' := r-d \left({\left({x, r}\right), \left({z, s}\right)}\right)$

and
 * $V := B_{r'}\left({z, s}\right) \cap S$

By definition of open ball:
 * $d \left({\left({x, r}\right), \left({z, s}\right)}\right) < r$

Then $r' > 0$

By $(2)$:
 * $V \in \mathcal B \left({z, s}\right)$

By proof of Open Ball of Point Inside Open Ball:
 * $B_{r'}\left({z, s}\right) \subseteq B_{r'}\left({x, r}\right)$

By Intersection is Subset:
 * $V = B_{r'}\left({z, s}\right) \cap S \subseteq B_{r'}\left({z, s}\right)$

By Set is Subset of Union:
 * $B_{r'}\left({x, r}\right) \subseteq U$

Thus by Subset Relation is Transitive: $V \subseteq U$

Ad $(N3)$:

Let $\left({x, y}\right) \in S, U_1, U_2 \in \mathcal B \left({x, y}\right)$.

In a case when $y > 0$ by $(2)$:
 * $\exists r_1 > 0: U_1 = B_{r_1}\left({x, y}\right) \cap S$

and
 * $\exists r_2 > 0: U_2 = B_{r_2}\left({x, y}\right) \cap S$

By Trichotomy Law for Real Numbers:
 * $r_1 \leq r_2$ or $r_1 \geq r_2$

WLOG: suppose
 * $r_1 \leq r_2$

Then by Open Ball is Subset of Open Ball:
 * $B_{r_1}\left({x, y}\right) \subseteq B_{r_2}\left({x, y}\right)$

By Set Intersection Preserves Subsets/Corollary:
 * $U_1 \subseteq U_2$

By Intersection with Subset is Subset:
 * $U_1 = U_1 \cap U_2$

Thus by Set is Subset of Itself:
 * $U_1 \subseteq U_1 \cap U_2$

In a case when $y = 0$ by $(3)$:
 * $\exists r_1 > 0: U_1 = B_{r_1}\left({x, r_1}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

and
 * $\exists r_2 > 0: U_2 = B_{r_2}\left({x, r_2}\right) \cup \left\{{\left({x, 0}\right)}\right\}$

By Trichotomy Law for Real Numbers:
 * $r_1 \leq r_2$ or $r_1 \geq r_2$

WLOG: suppose
 * $r_1 \leq r_2$

Then by Open Ball is Subset of Open Ball:
 * $B_{r_1}\left({x, r_1}\right) \subseteq B_{r_2}\left({x, r_2}\right)$

By Set Union Preserves Subsets/Corollary:
 * $U_1 \subseteq U_2$

By Intersection with Subset is Subset:
 * $U_1 = U_1 \cap U_2$

Thus by Set is Subset of Itself:
 * $U_1 \subseteq U_1 \cap U_2$

Thus by $(N0)$-$(N3)$, $(4)$ and Topology Defined by Neighborhood System:
 * $T = \left({S, \tau}\right)$ is a topological space.