Definition: Let $F \subseteq C(X,Y)$ where $X$ is a topological space, $(Y,d)$ a metric space. Then $F$ is called equicontinuous in $x$ iff
$$\forall \epsilon > 0: \exists V \in \mathcal{V}(x): \forall f \in F: f(V) \subseteq B_d(f(x), \epsilon)$$
Let $F = \{f_1, \dots, f_n\} \subseteq C(X,Y)$ be a finite collection of continuous functions where $X$ is an arbitrary topological space and $(Y,d)$ is a metric space. Prove that $F$ is equicontinuous.
Attempt:
Let $x \in X$ be fixed, $\epsilon > 0$. Put $V:= \bigcap_{j=1}^n f_j^{-1}(B_d(f(x), \epsilon))$. Then $V$ is a nbh of $x$, as $f_i$ is continuous and finite intersections of nbh's are nbh's.
Then, if $i \in \{1, \dots n\}$, it follows that $f_i(V) \subseteq B_d(f_i(x), \epsilon)$
Hence, $F$ is equicontinuous in $x$, and since $x$ was arbitrary, the result follows.
My problem with this solution is that $V$ seems to depend on $F$, and this isn't allowed. Any comments on this?
Why do you say $V$ can't depend on $F$? $V$ can and must depend on $F$. What wouldn't be allowed would be for $V$ to depend on $f_i\in F$, say. That $F$ is equicontinuous means that for the specific set of functions $F$, you can find a neighborhood on which each of the functions varies by less than $\epsilon$. Of course you can't find a neighborhood which would work no matter what the set $F$ is, otherwise you'd have a neighborhood $V$ on which every continuous function varies by at most $\epsilon$. That can't exist except in very simple topological spaces.
Your proof is correct.