Recall that if $X$ is a topological space, and $x\in X$ we say that $X$ is locally connected at $x$ if for every open set $V$ containing $x$ there exists a connected, open set $U$ with $x \in U \subseteq V$. The space $X$ is said to be locally connected if it is locally connected at $x$ for all $x$ in $X$.
Now let $X$ be a compact (is not necessarily Hausdorff) topological space and $A$ and $B$ two closed and locally connected subspaces of $X$. I am looking for conditions (especially equivalence condition ) under which $A\cup B$ to be locally connected subspace of $X$.
Best Answer
This is always true, and here is a proof.
For any point $x\in A$ which is not in $B$, we have that $A\cap B^c$ is an open neighborhood of $x$ in $A$ which does not intersect $B$. In particular, this means every open neighborhood of $x$ has a subneighborhood of $x$ which is contained entirely within $A$. By an application of the definition of locally connected to this subneighborhood, we get the required connected subneighborhood for every neighborhood of $x$ inside $A\cup B$. By interchanging $A$ and $B$, all we have to do now is to consider points inside $A\cap B$.
Let $x\in A\cap B$ and pick an open neighborhood $U\subset A\cup B$ of $x$. Up to shrinking $U$, we may assume $U$ has connected intersection with $A$. (If not, then we can find an open neighborhood of $x\in A$ which is a connected subneighborhood of $U\cap A$, and by the properties of the subspace topology we can find an open subset of $A\cup B$ which when intersected with $A$ gives this subneighborhood. By intersecting this subneighborhood with $U$, we get the required shrinking of $U$.)
Now for every point $p\in U\cap A\cap B$, let $V_{B,p} \subset B$ be an open connected neighborhood of $p$ in $B$ contained in $U\cap B$ (this is guaranteed from the fact that $B$ is locally connected). Let $V_B=\bigcup_{p\in U\cap A\cap B} V_{B,p}$, and pick some open $V\subset A\cup B$ which is open and has $V\cap B= V_B$, again guaranteed by the subspace topology. Up to replacing $V$ by $V\cap U$, we may assume that $V\subset U$.
Now I claim that $V\cup (U\cap (A\setminus B))$ is the required open connected neighborhood of $x$. We have $x\in V$ by construction, and as each of $V,U$, and $A\setminus B$ are open, we have that $V\cup (U\cap (A\setminus B))$ is open. If we can show $V\cup (U\cap (A\setminus B)) = (V\cap B) \cup (U\cap A)$, then we'll have connectedness: as $V\cap B=\bigcup_{p\in U\cap A\cap B} V_{B,p}$ and each $V_{B,p}\subset V\cap B$ meets $U\cap A$ in some point of $U\cap A\cap B$ by construction, we have that $(V\cap B)\cup (U\cap A)$ is connected.
Seeing that $V\cup (U\cap (A\setminus B))\subset (V\cap B)\cup (U\cap A)$ is straightforwards. If $p\in (U\cap (A\setminus B))$, then $p\in U\cap A$. If $p\notin (U\cap (A\setminus B))$, then it must be in $V$, and as $V\subset U$, we must have $p\notin (A\setminus B)$, or $p\in B$. So $p\in V\cap B$, and either way, $p\in (V\cap B)\cup (U\cap A)$.
Seeing that $(V\cap B)\cup (U\cap A) \subset V\cup (U\cap (A\setminus B))$ requires a little more care. If a point $p\in (V\cap B)\cup (U\cap A)$ is in $V\cap B$, then $p\in V$. If a point $p$ is not in $V\cap B$, then it must be in $(U\cap A)$, and either it's not in $B$, or it's in $B$ and not in $V$. In the first case, $p\in (U\cap (A\setminus B))$. The second case can't happen: the assumption that $p\in U\cap A$ and $p\in B$ forces $p\in U\cap A\cap B$, and by construction, $V$ contains all points of $U\cap A\cap B$. So $p\in V\cup (U\cap (A\setminus B))$, and this shows the equality of $V\cup (U\cap (A\setminus B))$ and $(V\cap B)\cup (U\cap A)$, so we're done.
The key difference between this and the link you posted in the comments is that the requirement that $A,B$ are both closed forces better behavior.