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Let $X,Y$ be topological spaces and $f:X \times Y \to \mathbb R$ continuous. It's from this answer that

If $Y$ is compact compact, then $$ X \to \mathbb R, x \mapsto \inf_{y\in Y} f(x,y) $$ is continuous.

Now $Y$ is not necessarily compact. Is $$ \phi: X \to \mathbb R \cup \{-\infty\}, x \mapsto \inf_{y\in Y} f(x,y) $$ lower (or upper) semi-continuous? In below attempt, I'm stuck at showing the existence of such $y_0$.

Clearly, $\operatorname{epi} f := \{(x, y, \lambda) \mid f (x,y) \le \lambda\}$ is closed. Our goal is to show that $\operatorname{epi} \phi$ is closed. Assume $(x_n, \lambda_n)$ is a sequence in $\operatorname{epi} \phi$ such that $x_n \to x\in X$ and $\lambda_n \to \lambda \in \mathbb R$. We want to show that $(x,\lambda) \in \operatorname{epi} \phi$.

Fix $\epsilon>0$. For each $n$, there is $y_n \in Y$ such that $f (x_n, y_n) \le \lambda_n +\epsilon$. If there exist $\color{blue}{y_0}\in Y$ and a subsequence $(y_{n_m})_m$ such that $y_{n_m} \to y_0$ as $m \to \infty$, then by l.s.c. of $f$ we get $f (x,y_0) \le \lambda+\varepsilon$. The claims then follows

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We will show that $\phi$ is upper semi-continuous (u.s.c). Fix $a \in \mathbb R$, we have $$ \{x: \phi (x) <a\}= \pi_X \left [\{(x,y): f(x,y) <a\} \right ] $$ is open because $f$ is continuous and the projection map $\pi_X$ is open. Then $\{x: \phi (x) \ge a\}$ is closed and thus $\phi$ is u.s.c.

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