Metric Space is T5

Theorem
Let $M = \left({X, d}\right)$ be a metric space.

Then $M$ is a $T_5$ space.

Proof
Let $A, B \subseteq X$ such that $A$ and $B$ are separated in $X$.

Then:
 * each point $x \in A$ has a neighborhood $N_{\epsilon_x} \left({x}\right)$ which is disjoint from $B$


 * each point $y \in B$ has a neighborhood $N_{\epsilon_y} \left({y}\right)$ which is disjoint from $A$.

Then:
 * $U_A = \displaystyle \bigcup_{x \in A} N_{\epsilon_x / 2} \left({x}\right)$


 * $U_B = \displaystyle \bigcup_{y \in B} N_{\epsilon_y / 2} \left({y}\right)$

are disjoint open neighborhoods of $A$ and $B$ respectively.

Hence the result by the definition of $T_5$ space.