Definition: Let $X$ be a space, a neighborhood of a subset $A$ of $X$ is a set $N$ containing an open set (in $X$) that contains $A$.
That is, a neighborhood need not be open.
Assume $Y'$ is a locally compact, $T_2$ space and that $X$ is locally connected.
Consider a continuous map $f:X \times Y\rightarrow X\times Y'$. Assume
(1). $\pi_1(f(x,y))=x$ for all $(x,y)\in X\times Y$. Where $\pi_1:X\times Y'\rightarrow X$ is the projection map.
(2). Further assume that for each $x\in X$, the map $\varphi_x:Y\rightarrow Y'$, given by $\varphi_x(y)=\pi_2(f(x,y))$, is an open map that is a homeomorphism onto its image.
Here $\pi_2: X\times Y'\rightarrow Y'$ is the projection map.
Further assume $(a,x)\in X\times U$, where $U$ is a precompact open subset of $x$ in $Y$
Note, $U$ is locally compact and Hausdorff $(\star)$
As $U$ is locally compact and Hausdorff, we can find precompact $V\owns x$ in $U$. Since $V$ is locally compact, we can find a precompact $W\owns x$ in $V$. Then, $\overline{W}\subseteq int(\overline{V})$.
Lemma: There exists a connected neighborhood $N_a$ of $a$ in $X$ such that for all $b\in N_a$, $\varphi_b(\overline{W})\subseteq \varphi_{a}(int(\overline{V}))$.
I am unable to prove the Lemma.
Comments:
[1]I tried looking the the fact that $f^{-1}(a \times \varphi_a(\overline{W}))$ is closed and contained in $X\times \overline{W}$ (by $(1)$). However, I have gotten nowhere.
[2]. $\star$ follows from the fact that each $\phi_{x}$ is a homeomorphism onto image.
[3]. $Y$ is Hausdorff and locally compact since $\phi_{x}$ is a homeomorphism onto image and is open.
[4]. We need to find connected $N_a$ of $a$ such that for all $b\in N_a$, $\overline{W}\subseteq \phi_{b}^{-1}(\phi_{a}(int(\overline{V}))$.
As far as I understand the question, $V$ is a neighborhood of $x$, hence we can replace it by its interior and simply suppose $V$ is open. Then $V\subset$int$(\overline V)$ and we seek the inclusion $\varphi_b(\overline W)\subset\varphi_a(V)$. Then $\varphi_a(V)$ is open in $Y'$ and $f^{-1}(X\times\varphi_a(V))$ is open in $X\times Y$. Since $$ \{a\}\times\overline W\subset f^{-1}(X\times\varphi_a(V)), $$ and the latter set is open neighborhood in $X\times Y$ of the compact set $\{a\}\times\overline W$. By the general tube lemma, there is an open neighborhood $A$ of $a$ such that $$ A\times\overline W\subset f^{-1}(X\times\varphi_a(V)). $$ This implies that for every $b\in A$ the condition we seek holds. But $X$ is locally connected, and we can choose a smaller connected $A$.