Your suggestion would take care of the long line, since it is a connected linearly ordered space, but it wouldn’t help with the topologist’s sine curve for instance. A connected linearly ordered space with endpoints is compact, so the continuous image of such a space is also compact and connected, but no compact, connected subset of the topologist’s sine curve contains the origin and at least one other point of the curve.
It also won’t help with any of the various countably infinite connected Hausdorff spaces that have been constructed. A generalized path in such a space would be a countable, connected, compact Hausdorff space, and no such space exists.
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.
Best Answer
There are no such connections.