Equivalence of Definitions of Weakly Locally Connected at Point/Definition 2 implies Definition 1

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $x \in S$. Let every open neighborhood $U$ of $x$ contain an open neighborhood $V$ such that every two points of $V$ lie in some connected subset of $U$.

Then:
 * $x$ has a neighborhood basis consisting of connected sets.

Proof
Let $\mathcal B = \set{B \subseteq S: B \text{ is a connected neighborhood of } x}$

It will be shown that $\mathcal B$ is a neighborhood basis consisting of connected sets.

Let $N$ be any neighborhood of $x$.

By definition of a neighborhood there exists an open neighborhood $U$ of $x$ such that $U \subseteq N$

By assumption there exists an open neighborhood $V$ of $x$ such that every two points of $V$ lie in some connected subset of $U$.

For each $y \in V, y \neq x$, let $C_y$ be a connected subset of $U$ such that $x,y \in C_y \subseteq U$.

Let $C = \set {x} \cup \bigcup \set {C_y:y \in V, y \neq x}$

From Every Singleton is Connected in Topological Space, $\set {x}$ is connected.

From Union of Connected Sets with Common Point is Connected, $C$ is a connected subset.

From Union is Smallest Superset, $C \subseteq U \subseteq N$.

Now $x \in C$ and for each $y \in V, y \neq x$ then $y \in C_y \subseteq C$.

Hence $V \subseteq C$.

By definition of a neighborhood then $C$ is a connected neighborhood of $x$.

That is, $C \in \mathcal B$.

Since $N$ was an arbitrary neighborhood of $x$ then $\mathcal B$ is a neighborhood basis of $x$ consisting of connected sets.