Equivalence of Definitions of Locally Connected Space/Definition 2 implies Definition 1

Theorem
Let $T = \struct{S, \tau}$ be a topological space. Let each point of $T$ have a neighborhood basis consisting of connected sets of $T$.

Then
 * each point of $T$ has a local basis consisting entirely of connected sets in $T$.

Proof
Let $x \in S$ and $x \in U \in \tau$.

Let $\mathcal B_x = \set{ W \in \tau : x \in W, W \text{ is connected in } T}$. By definition of local basis, we have to show that there exists a connected open set $V \in \mathcal B_x$ with $x \in V \subset U$.

Let $V = \map {\operatorname{Comp}_x} U$ denote the component of $x$ in $U$.

From Open Set in Open Subspace, it suffices to show that $V$ is open in $U$.

By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points.

Let $y \in V$.

By assumption, there exists a connected neighborhood $W$ of $y$ in $T$ with $W \subset U$.

By Neighborhood in Topological Subspace, $W$ is a neighborhood of $y$ in $U$.

By definition of component:
 * $W \subseteq \map {\operatorname{Comp}_y} U = \map {\operatorname{Comp}_x} U = V$

By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$.

Because $y$ was arbitrary, $V$ is open in $U$.

From Open Set in Open Subspace, $V$ is open in $T$.

Hence $V \in \mathcal B_x$.

Because $U$ was arbitrary, $\mathcal B_x$ is a local basis of $x$ consisting of (open) connected sets.

Since $x$ was arbitrary, then each point of $T$ has a local basis consisting entirely of connected sets in $T$.