$f\circ g$ continuous, $f$ local homeomorphism, $g$ continuous in a different topology $\implies g$ is continuous I've asked this question before but neglected some assumptions and got a less than useful answer as a result, so I'm going to try again.

Let $g:I\times I\to Y$ (where $I=[0,1]$) be a function such that $g(0,-):I\to Y$ and $g(-,x):I\to Y$ for each $x\in I$ are continuous, $f:Y\to Z$ be a local homeomorphism, and suppose $h=f\circ g$ is a continuous function. Show that $g$ is continuous.

The condition on $g$ can also be understood as the statement that $g$ is continuous in the comb-like topology on $I\times I$ generated by open sets of the form $\{0\}\times U$ or $U\times\{x\}$ where $U$ is open in $I$. Importantly (I think), this topology is connected (although not locally connected).
The statement is "intuitively obvious" but I'm having trouble with the proof. Here is a partial proof assuming the following lemma, which basically "localizes" the original statement:

Lemma (unproven): For any $(x,y)$ there is a neighborhood $U\times V\ni(x,y)$ such that if there is a $x'\in U$ such that $g$ is continuous in the subspace topology on $\{x'\}\times V$ (or alternatively, $y'\in V\mapsto g(x',y')$ is continuous), then $g$ is continuous on $U\times V$.

Let $C=\{(x,y)\mid g\mbox{ is continuous at }(x,y)\}$ and $A=\{x\in I:\{x\}\times I\subseteq C\}$. It is sufficient to prove that $A=I$, and by the Lemma, $g(0,-)$ is continuous implies $0\in A$. We will prove that $A$ is clopen; from this it follows that $A=I$ since $I$ is connected. Let $x\in I$. We wish to show that there is a neighborhood $U\ni x$ such that for all $x'\in U$, $x'\in A\leftrightarrow x\in A$.
Define $S_x=\{(x',y)\mid\exists V\ni y\,(\{x\}\times V\subseteq C\leftrightarrow\{x'\}\times V\subseteq C)\}$. Then $S_x^\circ$ is an open set, so by the tube lemma (using compactness of $I$), if $\{x\}\times I\subseteq S_x^\circ$ then there is a neighborhood $U$ such that $U\times I\subseteq S_x^\circ\subseteq S_x$, and then for all $x'\in U$, the definition of $S_x$ implies $x'\in A\leftrightarrow x\in A$ as desired.
It remains to prove that $\{x\}\times I\subseteq S_x^\circ$. Let $y\in I$; then by the Lemma there is a neighborhood $(x,y)\in U\times V$ with the property that if $\{x'\}\times V\subseteq C$ (which is slightly stronger than continuity of $y'\in V\mapsto g(x',y')$) then $U\times V\subseteq C$; to see that $U\times V\subseteq S_x$ note that if $\{x'\}\times V\subseteq C$ then $\{x\}\times V\subseteq C$ and vice versa because each one implies $U\times V\subseteq C$.
 A: As it turns out, the Lemma, and hence the full statement, is indeed true under the following additional assumption:

Around every point $y\in Y$ there is an open neighborhood $W$ such that $f:W\to f[W]$ is a homeomorphism, $f[W]$ is open, and $W$ is closed in $f^{-1}[f[W]]$.

(This is provable if $f$ is a covering map, and might even be equivalent to $f$ being a covering map, I'm not sure.) The proof is completed along the lines of the OP. The key step is the following fact about connected topologies:

$(*)$ If $f:X\to Y$ is continuous and $X$ is connected, $U\subseteq Y$ is clopen and for some $x$, $f(x)\in U$, then $f:X\to U$.

The proof is a simple consequence of the connectedness of $f[X]$.
Turning to the lemma, given $(x,y)$ we have $g(x,y)\in Y$, so there is a neighborhood $W\ni g(x,y)$ on which $f$ is a homeomorphism, and then by continuity of $h$ there is a neighborhood $U\times V$ such that $h[U\times V]\subseteq f[W]$. Additionally (using local connectedness of $I$ or by choosing $U$ and $V$ to be open intervals), we can ensure that $U$ and $V$ are connected. Now suppose that $x'$ is given such that $g\restriction\{x'\}\times V$ is continuous; we wish to prove that $g\restriction U\times V$ is continuous.
First, we prove $g[U\times V]\subseteq W$. Note that $g[U\times V]\subseteq f^{-1}[f[W]]$ because $f[g[U\times V]]=h[U\times V]\subseteq f[W]$. Let $(z,w)\in U\times V$. 


*

*Since $g\restriction U\times \{y\}:U\times \{y\}\to f^{-1}[f[W]]$ is continuous and $g(x,y)\in W$, by $(*)$ we also have $g(x',y)\in W$.

*Since $g\restriction \{x'\}\times V:\{x'\}\times V\to f^{-1}[f[W]]$ is continuous and $g(x',y)\in W$, by $(*)$ we also have $g(x',w)\in W$.

*Since $g\restriction U\times \{w\}:U\times \{w\}\to f^{-1}[f[W]]$ is continuous and $g(x',w)\in W$, by $(*)$ we also have $g(z,w)\in W$.


Since $z,w$ were arbitrary, this shows $g[U\times V]\subseteq W$. (We could also have shown, as suggested in the OP, that the topology generated by $\{x'\}\times V'$ and $U'\times\{y\}$ for each $U'$ open in $U$, $V'$ open in $V$, $y\in V$ is connected, and then gotten by with just one application of $(*)$.)
The rest is straightforward. Since $g\restriction U\times V:U\times V\to W$ it is sufficient to prove that $g$ is continuous in the subspace topology on $W$, so that for every open $W'\subseteq W$, $(g\restriction U\times V)^{-1}[W']$ is open in $U\times V$:
\begin{align}
(g\restriction U\times V)^{-1}[W']&=(g\restriction U\times V)^{-1}[(f\restriction W)^{-1}[(f\restriction W)[W']]]\\
&=((f\restriction W)\circ(g\restriction U\times V))^{-1}[(f\restriction W)[W']]\\
&=(f\circ(g\restriction U\times V))^{-1}[(f\restriction W)[W']]\\
&=((f\circ g)\restriction U\times V)^{-1}[(f\restriction W)[W']]\\
&=(h\restriction U\times V)^{-1}[(f\restriction W)[W']],
\end{align}
which is manifestly open.
