Let $X$ be compact, $\mathcal{F}$ be a family of equicontinuous real valued functions on $X$. Define $A:=\{a\in X|\{f(a): f\in\mathcal{F}\}\text{ is bounded}\}$. Show that $A$ is both closed and open.
Attempt: Let $a\in A$. By equicontinuity on compact set we have uniform equicontinuity. Since $f(a)$ is bounded for all $f\in\mathcal{F}$, then there exists $\delta>0$, such that if $d(a, x)<\epsilon$ then $|f(a)-f(x)|<\epsilon$. Thus, $f(x)$ are also bounded if $x$ is in the $\delta$-neighborhood of $a$, so it follows that $A$ is open.
Let $l$ be a limit point of $A$, then there exists a sequence of $\{a_n\}\in A$ such that $a_n\rightarrow l$. By continuity of each function $f\in\mathcal{F}$, $\{f(a_n)\}$ is a bounded sequence that converges to $f(l)$. By equicontinuity, for all $a\in A$ within $\delta_l$-neighborhood of $l$, $|f(a)-f(l)|<\epsilon$ for all $f\in\mathcal{F}$. Therefore $f(l)$ is bounded, so $l$ is contained in $A$.
Question: is the above correct? I only used compactness for uniform equicontinuity.
Let $a \in A$. So there is some $B$ such that all $|f(a)| \le B$ for all $f \in \mathcal{F}$. Let $\delta >0$ be the (uniform) equicontinuity constant for $\varepsilon =1$, so that
$$d(x,a) < \delta \implies \forall f \in \mathcal{F} |f(x) - f(a)| < 1\text{.}$$
But then for all $x \in B(a,\delta)$ we have that $|f(x)| \le |f(x) - f(a)| + |f(a)| \le 1+B$ for all $f \in \mathcal{F}$, so that $B(a, \delta) \subseteq A$ and $a$ is an interior point. This is essentially what you did as well, it's better to choose an explicit $\varepsilon$ to be more concrete, I think.
If $a \notin A$, $\{f(a): f \in \mathcal{F}\}$ is not bounded, so for all $N \in \mathbb{N}$ we can find some $f$ such that $|f(a)| > N$.
Take the same $\delta$ for $\epsilon=1$ as before. If $x \in B(a,\delta)$, let $N \in \mathbb{N}$. For some $f$ ,$|f(a)| > N+1$ and so $|f(x)| > N$ (as otherwise for this $f$, $|f(a)| \le |f(a) -f(x)| + |f(x)| < N+1$ which contradicts how we chose $f$). So $B(a,\delta) \subseteq X\setminus A$ and the latter set is open, and $A$ is closed.
No need for limit points in part 2, just a symmetric argument: if you uniformly stay near a function with (un)bounded images, you belong to the same class. A metric on $X$ (which you use, also in your limit point argument), is not given, but I used it to stay close to your attempts. It suffices to just have open neighbourhoods of $a$ throughout. Equicontinuity is defined for all spaces $X$; only the image needs to be a uniform space (or a metric space, as here). Also, note that we used only standard (not uniform) equicontinuity at $a$ for both parts. So the compactness of $X$ plays no rôle at all, as far as I can see.