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

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

Let $x \in S$. Let $x$ have a neighborhood basis consisting of connected sets.

Then:
 * Every open neighborhood $U$ of $x$ contains an open neighborhood $V$ such that every two points of $V$ lie in some connected subset of $U$.

Proof
Let $U$ be an open neighborhood of $x$.

By assumption there exists a connected neighborhood $C$ of $x$ such that $C \subseteq U$.

By definition of a neighborhood there exists an open neighborhood $V$ of $x$ such that $V \subseteq C$.

From Subset Relation is Transitive, $V \subseteq U$.

By definition of a subset, for any two points $y,z \in V$, $y,z \in C$.

The result follows.