Let $X$ be a (metric) space. Let $S$ and $L_i$ ($i\in I$) be connected subsets of $X$. Assume that $S\cap L_i \neq \phi$. Show that $S\cup (\cup_i L_i)$ is a connected subset of $X$.
My work: I know that union of two connected set is connected if the intersection is non-empty. Using this fact, it is easy to see that $S\cup L_i$ is a connected subset for all $i$. But I can not proceed from here because I was trying to use that fact that if $x,y\in S\cup ( \cup_i L_i)$, then there exists some connected set $A$ such that $x,y\in A$. Now case 1: $x,y\in$S . Case 2: WLG if $x\in$S and $y\in$$L_i$ for some i and case 3 : $x,y\in L_i$ for some $i$ for this case, I see that x,y can lie in the same connected subset but what is for the case if $x\in L_i$ and $y\in L_j$ for $i\neq j$. Because it is not given that $L_i$'s are disjoint or not. Help me to understand this.
You're on the right track, you just need to use induction to show that the whole union is empty. If $I$ is not assumed to be countable, this may be a bit awkward, if you're not comfortable with transfinite induction. So another way is to argue as follows:
Set $Y=S\cup \bigcup_i L_i$, let $s_0$ be some point in $S$, and let $R$ be the connected component of $Y$ which contains $s_0$. Since $S$ is connected, necessarily $R$ contains all of $S$. Let $i_0\in I$; we know $L_{i_0}$ intersects $S$, so it intersects $R$, so (since $L_{i_0}$ is connected and $R$ is a connected component of $Y$) necessarily $R$ contains all of $L_{i_0}$. Since $i_0$ was an arbitrary element of $I$, we get that $R$ contains $L_i$ for every $i$. Hence $R$ contains $S\cup \bigcup_i L_i=Y$. But $R$ was a connected component of $Y$, so it is contained in $Y$ as well, which means $Y=R$. By definition, $R$ is connected, so $Y=R$ is connected, as needed.