I thought about applying the proof for open functions, namely:
Proposition: If $f$ is a continuous open function from a locally connected space $X, \tau$ onto a space $T, \tau'$, then $Y$ is locally connected.
Proof: Suppose $y \in Y$ and $U$ is any neighborhood of $y$. Since $f$ is onto, there is $x \in X$ such that $f(x) = y$. Then $f^{-1}(U)$ is a neighborhood of $x$. Since $X$ is locally connected, there is a connected neighborhood $V$ of $x$. Since $f$ is both continuous and open, $f(V)$ is a connected neighborhood of $y$ with $f(V) \subset U$. Therefore $Y$ is locally connected.
I want to apply this somehow, but I can't figure out a way how.
I think that we could try to apply components to this. Components are closed, but if we have local connectedness then we know they're clopen, which means we might be able to travel along the function through the open attribute, and then claim something about being closed. That being said, the problem is we assume we have a continuous closed function. Yeah, pretty stuck on this.
Best Answer
We can unify the cases for closed / open surjective maps in a single proof: they are quotient maps, i.e. $f^{-1}[O]$ is open in $X$ iff $O$ is open in $Y$, for all $O \subset Y$.
Standard fact: a space $X$ is locally connected iff the components of every open subspace $O$ are open (in $O$ or $X$, that's the same). This is in all the textbooks.
Now if $O \subset Y$ is open and $C \subset O$ is a (connected) component of $O$, then check that every component of $f^{-1}[C]$ is a component of $f^{-1}[O]$, using continuity of $f$. Next, $f^{-1}[O]$ is open in $X$, so by local connectedness of $X$ we know all components of $f^{-1}[C]$ are open in it, and so $f^{-1}[C]$ is open. But then $f$ being quotient gives that $C$ is open in $O$ and you are done.
So we get a more general proof using the notion of quotient maps.