By a corollary of the Arzela-Ascoli theorem, I think we just need to show that $F$ is compact. But I'm unsure how to go about doing this.
p.s. Question taken from a past-year exam paper.
p.p.s. "Equicontinuous" in the question means uniformly equicontinuous, as this is the terminology we have adopted in the textbook and in class.
Because the convergance is uniform $f_n\to f$ we have that $f$ is continuous on $X$. Let $x_0\in X$ and $\varepsilon>0$. Then there is a $k_0\in \Bbb N:$ for every $x\in X,k>k_0\Rightarrow |f_k(x)-f(x)|<\varepsilon/3$.
Also there is a $\delta_0>0:x\in X,|x-x_0|<\delta_0\Rightarrow |f(x)-f(x_0)|<\varepsilon/3$.
Also due to the continuity of $f_k$ we have that for every $k=1,2,\dots,k_0 $ there is a $\delta_k>0:x\in X,|x-x_0|<\delta_k\Rightarrow |f_k(x)-f_k(x_0)|<\varepsilon$.
Let $\delta=\min\{\delta_0,\delta_1,\dots,\delta_k\}$. Then for every $k\in \Bbb N$ we have $x\in X,|x-x_0|<\delta\Rightarrow |f_k(x)-f_k(x_0)|<\varepsilon$.