Continuity of max (moving the domain and the function)

Note that the set $C:=\{(x,t)\mid x\in\Bbb R,\ t\in[0,c(x)]\}$ is closed and contains the graph of $c$, the set $G(c) = \{(x,c(x))\mid x\in\Bbb R\}$, which is closed in $\Bbb R \times [0,1]$

Show that the preimage of an open subbase set $(-\infty,a)$ under $F$ is open: This is the set $\Bbb R-\{x\mid \exists t\le c(x),f(x,t)\ge a\}$. The complement can be expressed as $\pi_{\Bbb R}(f^{-1}([a,\infty)\cap C)$. Since $[a,\infty)$ is closed, its preimage under $f$ is closed. But if we now apply the projection $\pi_{\Bbb R}$, we get a closed set again. This is because the projection $X\times Y\to X$ is closed if $Y$ is compact, so it is in the case $\Bbb R\times[0,1]\to\Bbb R$.

Still, there is a problem if we want to apply the same argument to an open subbase set $(a,\infty)$ since the restriction of an open map, like the projection, to a closed subset isn't necessarily open. In this case, however, $\pi_{\Bbb R}|_C$ is indeed open. This is obvious on $C\setminus G(c)$. And on $G(c)$, we have $\pi_{\Bbb R}((U×V)\cap G(c))=c^{-1}(V)\cap U$.