Equivalence of Definitions of Locally Connected Space

$(1)$ implies $(2)$
Let $T = \struct{S, \tau}$ be locally connected by Definition 1:
 * each point of $T$ has a local basis consisting entirely of connected sets in $T$.

From Local Basis for Open Sets Implies Neighborhood Basis of Open Sets, it follows directly that:
 * each point of $T$ has a neighborhood basis consisting entirely of connected sets in $T$.

That is, $T = \struct{S, \tau}$ is locally connected by Definition 2.

$(2)$ implies $(1)$
Let $T = \struct{S, \tau}$ be locally connected by Definition 2:
 * each point of $T$ has a neighborhood basis consisting of connected sets of $S$ in $T$.

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

By definition of local basis, we have to show that there exists a connected open set $V \in \tau$ with $x \in V \subset U$.

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

By definition of topological 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 $T$.

Because $U$ was arbitrary, $x$ has a local basis of (open) connected sets.

That is, $T = \struct{S, \tau}$ is locally connected by Definition 1.

$(1)$ implies $(3)$
Let $T = \struct{S, \tau}$ be locally connected by Definition 1:
 * each point $x$ of $T$ has a local basis $\mathcal D_x$ consisting entirely of connected sets in $T$.

By definition of local basis, each of these connected sets in $\mathcal D_x$ is open in $T$.

Consider the set $\displaystyle \mathcal D = \bigcup_{x \mathop \in S} \mathcal D_x$.

From Union of Local Bases is Basis, $\mathcal D$ is a basis for the topology $\tau$.

Since each $\mathcal D_x$ consists entirely of connected sets, $\mathcal D$ also consists entirely of connected sets.

That is, $T = \struct{S, \tau}$ is locally connected by Definition 3.

$(3)$ implies $(1)$
Let $T = \struct{S, \tau}$ be locally connected by Definition 3:
 * $T$ has a basis consisting of connected sets in $T$.

For each $x \in S$ we define:
 * $\mathcal B_x = \left\{{B \in \mathcal B: x \in B}\right\}$

From Basis induces Local Basis, $\mathcal B_x$ is a local basis.

As each element of $\mathcal B_x$ is also an element of $\mathcal B$, it follows that $\mathcal B_x$ is also formed of connected sets.

Thus, for each point $x \in S$, there is a local basis which consists entirely of connected sets.

That is, $T = \struct{S, \tau}$ is locally connected by Definition 1.

$(3)$ implies $(4)$
Let $T = \struct{S, \tau}$ be locally connected by Definition 3:
 * $T$ has a basis consisting of connected sets in $T$.

Let $U$ be an open subset of $T$.

From Open Set is Union of Elements of Basis, $U$ is a union of open connected sets in $T$.

From Open Set in Open Subspace, $U$ is a union of open connected sets in $U$.

From Components are Open iff Union of Open Connected Sets, the components of $U$ are open in $U$.

From Open Set in Open Subspace then the components of $U$ are open in $T$.

That is, $T = \struct{S, \tau}$ is locally connected by Definition 4.

$(4)$ implies $(3)$
Let $T = \struct{S, \tau}$ be locally connected by Definition 4:
 * the components of the open sets of $T$ are also open in $T$.

That is, $T = \struct{S, \tau}$ is locally connected by Definition 3.

Also see

 * Equivalence of Definitions of Locally Path-Connected Space, whose proof is almost the same