Let $(X, \mathcal{T})$ be a topological space, $(Y,d)$ be a metric space, and $H \subseteq C(X,Y)$. Then, $\operatorname{cl}(H)$ is equicontinuous too in the pointwise topology (subspace on the product topology $Y^X$).
My attempt:
We want to prove that $\operatorname{cl}(H)$ is equicontinuous in $x \in X$/
Let $\epsilon > 0$. Choose $V \in \mathcal{V}_X(x)$ open such that for all $f \in H: f(V) \subseteq B_d(f(x), \epsilon)$.
Then, we have to find $W \in \mathcal{V}_X(x)$ such that $f(V) \subseteq B_d(f(x), \epsilon)$ $\forall f \in \operatorname{cl}(H)$.
How do I proceed?
Let $p \in X$ and $\varepsilon > 0$ be given. By equicontinuity, pick $V_p$, an open neighbourhood of $p$ such that
$$\forall h \in H: h[V_p] \subseteq B_d(h(p), \frac{\varepsilon}{3})$$
I claim that this $V_p$ also works for showing the equicontinuity of $\overline{H}$ (the pointwise closure of $H$ in $Y^X$) for $\varepsilon >0$, and to this end we let $f$ be any function in $\overline{H}$ and we I'll show that $$f[V_p] \subseteq B_d(f(p), \varepsilon)$$
and to this end, let $x \in V_p$ be arbitrary. (it's fixed for now.) Then define $$O = (\pi_x)^{-1}[B_d(f(x),\frac{\varepsilon}{3})] \cap (\pi_{p})^{-1}[B_d(f(p), \frac{\varepsilon}{3})]$$
which is an open subset of $Y^X$ in the product topology (which makes all $\pi_q, q \in X$ continuous,in particular $\pi_x$ and $\pi_p$, so $O$ is a finite intersection of two open sets of $Y^X$, hence open). By definition $f \in O$, and as $f \in \overline{H}$, there is some $h_0 \in H$ with $h_0 \in O$.
The last fact implies that $d(h_0(x), f(x)) < \frac{\varepsilon}{3}$ and also that $d(h_0(p), f(p)) < \frac{\varepsilon}{3}$ from the definitions. Also, $h_0 \in H$ implies that $$h_0[V_p] \subseteq B(h_0(p), \frac{\varepsilon}{3})$$
So: $$d(f(p),f(x)) < d(f(p), h_0(p)) + d(h_0(p), h_0(x)) + d(h_0(x), f(x)) < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$$
where the two outer inequalities are from $h_0 \in O$ and the middle from the equicontinuity and $x \in V_p$.
As $x \in V_p$ was arbitary, we have indeed shown $f[V_p] \subseteq B_d(f(p),\varepsilon)$, as was required.