Locally connected and connected Let X be locally connected, $A \subset X$ arbitrary. Let $S \subset A$ be connected and open in A. Show $S=U\cap A$ where $U$ is connected and open.
I think that I have solved already but I never used that X was locally connected so if you could help me a little bit....
Like $S$ is open in $A$ there exists $K\subset X$ open such that $S=K\cap A$. Let suppose K is disconnected in $X$. Thus there exist $J,L\subset X$ open  and  $K\cap J\neq \emptyset$ $\neq$ $K\cap L$ and $J\cap L \cap K= \emptyset$ such that $K=(J\cap K)\cup (L\cap K)$. Then we have $S=((J\cap K)\cup (L\cap K))\cap A=(J\cap K \cap A)\cup (L\cap K \cap A)$. So $S$ would be disconnected in $A$, hence $K$ is connected in $X$.
I rushed in the conclusion, let see....
If both $(J\cap K \cap A) =\emptyset= (L\cap K \cap A)$, then $U=\emptyset$ would do the job. So lets suppose $(J\cap K \cap A)\neq \emptyset$, like $J,K$ are open in X then $J\cap K$ is open in X, so $U=J\cap K$ its what we need. Not is not what we need, I do not know if $U=J\cap K$ is connected, so here I must use the X is locally connected....

As far as what you've done goes, I'd remark that

... since $J\cap K\cap A$ happens to be empty because I said so, there is no absurd here.

Rather, I think you should observe that connected components of open sets in a locally connected topological space are open, and then you should consider the connected component of $K$ that contains $S$.
