Fix $t\in\Bbb R$. We just have to show sequential continuity, since we are working in a metric space. Let $\{t_n\}\subset\Bbb R$ a sequence which converges to $t$. Since $X$ is compact, we can find $x_n$ such that $m(t_n)=f(x_n,t_n)$. We show that for each subsequence of $\{t_n\}$ we can find a further subsequence $\{t_{n_k}\}$ such that $m(t_{n_k})\to m(t)$. It will show that $m(t_n)\to m(t)$.
Let $\{t_{n'}\}$ a subsequence of $\{t_n\}$. The sequence $\{x_{n'}\}$ admits a converging subsequence $\{x_{n_k}\}$, say to $x$. Then $(t_{n_k},x_{n_k})\to (t,x)$ and we conclude using the continuity of $f$ (and the fact that $f(x_n,t_n)\geq f(y,t_n)$ for all $y\in Y$, to show that $f(x,t)=u(t)$.