Let $(X, d)$ be a compact metric space and let $\{f_{\alpha} : \alpha \in A\}$ be a uniformly bounded and equicontinuous family of functions on $X$. Define $$f(x) = \sup_{\alpha \in A} f_{\alpha}(x)$$ Then, for any $t \in \mathbb{R}$, the set $\{x \in X \ : \ f(x) < t \}$ is an open set in $X$.
My attempt: The set $$\{x \in X \ : \ f(x) < t \} \implies \cap_{\alpha}\{x \in X \ : \ f_{\alpha}(x) < t \} $$ As each $\{f_{\alpha}\}$ is continuous, so each set of the form $$\{x \in X \ : \ f_{\alpha}(x) < t \}$$ is open. But arbitrary intersection of open sets is not open. How to see? Where i am wrong.
You have not made use of the equicontinuity of the family ${\cal F}$.
Let a $t\in{\mathbb R}$ be given, and let $p\in X$ be an element of the set $S_t:=\{x\in X\,|\, f(x)<t\}$. Let $$\epsilon:={t-f(p)\over2}>0\ .$$ By equicontinuity of the $f_\alpha$ there is a $\delta>0$ such that $$f_\alpha(x)\leq f_\alpha(p)+\epsilon \leq f(p)+\epsilon \quad\forall x\in U_\delta(p), \>\forall\alpha\in A\ .$$ Consider now an arbitrary $x\in U_\delta(p)$. Then we can say that $$f(x):=\sup_{\alpha\in A} f_\alpha(x)\leq f(p)+\epsilon<t\ .$$ This shows that $U_\delta(p)\subset S$, and proves that $S_t$ is open.
It turns out that the compactness of $X$ was not needed.