Continuity of inf function

Let $X,Y$ be topological spaces, $Y$ is compact and $f:X\times Y\to \mathbb{R}$ be a continuous function. Define $$g(x)=\inf_{y\in Y}f(x,y)$$
Then $g$ is continuous.

$\textbf{Proof}$. It is enough to check that for all real $a$ the sets $g^{-1}((-\infty,a))$ and $g^{-1}((a,\infty))$ are both open in $X$.
First rewrite $$g^{-1}((-\infty,a))=\{x\in X\ | \ \inf_{y\in Y}f(x,y)<a\}=\{x\in X\ | \ \exists y\in Y, f(x,y)<a\}.$$
Observe that
$$\{x\in X\ | \ \exists y\in Y, f(x,y)<a\}=\bigcup_{y_0\in Y}\{x\in X\ | \ f(x,y_0)<a\}.$$ Since $f$ is continuous, the function $h_{y_0}(x)=f(x,y_0)$ is continuous as well. This means that the set $\{x\in X\ | \ f(x,y_0)<a\}$ is open (being preimage of the open interval $(-\infty,a)$ under the continuous function $h_{y_0}$). Hence $\{x\in X\ | \ \exists y\in Y, f(x,y)<a\}$ is a union of open sets - thus it is open.
Now $$g^{-1}((a,\infty))=\{x\in X\ | \ \inf_{y\in Y}f(x,y)>a\}.$$
Denote $$V_n=\left\{x\in X\ | \ \forall y\in Y,\ f(x,y)> a+\frac{1}{n}\right\}$$ so that $\displaystyle\bigcup_n V_n=\{x\in X\ | \ \inf_{y\in Y}f(x,y)>a\}$. Thus it suffices to prove that all of the $V_n$ are open. Fix $n$ and for brevity $V=V_n$ and $\displaystyle b=a+\frac{1}{n}$. We write
$$V=\{x\in X\ | \ \forall y\in Y,\ f(x,y)> b\}.$$
Assume $V$ is nonempty and let $\hat{x}\in V$. Since $f$ is continuous, for any $\bar{y}\in Y$ there are open neighbourhoods $U_\bar{y}$ of $\hat{x}$ and $W_\bar{y}$ of $\bar{y}$ such that for any $(x,y)\in U_\bar{y}\times W_\bar{y}$, $f(x,y)>b$. The sets $W_\bar{y}$ form an open cover of $Y$ and so we may choose $y_1,\ldots,y_n\in Y$ such that the sets $W_{y_i}$ also cover $Y$. Let $\displaystyle U=\bigcap_i U_{y_i}$. Thus $U$ is nonempty ($\hat{x}\in U$) open set. Let $x\in U$ and $y\in Y$. Thus $y\in W_{y_i}$ for some $i$, and so $(x,y)\in U_{y_i}\times W_{y_i}$ hence $f(x,y)>b$. Hence $x\in V$. This means that $U$ is an open neighbourhood of $\hat{x}$ which is contained in $V$. Thus $V$ is open.

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