Are these two definitions of equi-continuity equivalent? Let $(X,\mathcal{T}),(Y,\mathcal{U})$ be topological spaces and $F \subseteq C(X,Y)$. $\newcommand{\Ball}{\operatorname{Ball}}$

Topological equi-continuity:
$F$ is equi-continuous at $x \in X,y \in Y$ if for every open set $S \ni y$, there exist neighbourhoods $U$ of $x$ and $V$ of $y$ such that $\forall f \in F, f[U] \cap V \neq \emptyset \implies f[U] \subseteq S$.
$F$ is equi-continuous at $x \in X$ if it is equi-continuous at every $x,y$ for $y \in Y$.

Now suppose that $\mathcal{U}$ is induced by a metric $d$ on $Y$.

Metric equi-continuity:
$F$ is equi-continuous at $x \in X$ if for every $\epsilon > 0$, there exists a neighbourhood $U^\circ \ni x$ with $\forall f \in F, f[U] \subseteq \Ball(f(x);\epsilon)$.

Proof:
Topological $\implies$ Metric: Let $\epsilon > 0, x \in X$. Suppose that $F$ is equi-continuous at $x$. For every $y \in Y$, there exist neighbourhoods $U_y$ of $x$ and $V_y$ of $y$ with $\forall f \in F, f[U_y] \cap V_y \neq \emptyset \implies f[U_y] \subseteq \Ball(y,\epsilon)$. Substituting $y = f(x)$ gives $f[U_{f(x)}] \cap V_{f(x)} \neq \emptyset \implies f[U_{f(x)}] \subseteq \Ball(f(x),\epsilon)$. The former condition always holds since $x \in U_{f(x)}, f(x) \in V_{f(x)}$.
Now taking intersections of $U_{f(x)}$ might yield a set that might not be a neighbourhood of $x$. What can be done to solve this?
Metric $\implies$ Topological: Let $y \in Y$ and $S$ be an open set containing $y$. Since $d$ generates the topology on $Y$, there exists $\epsilon > 0$ with $\Ball(y;\epsilon) \subseteq S$. By metric equi-continuity, there exists $U^\circ \ni x$ with $f[U] \subseteq \Ball(f(x);\epsilon)$ for all $f$. For each $f$ with $f(x) \neq y$, there exists a neighbourhood $y \in \Ball(y;\delta_y)$, but if we take the intersection of all $\Ball(y;\delta_y)$, the result might not be a neighbourhood of $y$ anymore and the same problem arises.
 A: I’ll assume that $\Ball(y,\epsilon)=\{z\in Y:d(y,z)<\epsilon\}$  means an open ball.
Metric $\Rightarrow$  Topological: Let $y \in Y$ and $S$ be an open set containing $y$. Since $d$ generates the topology on $Y$, there exists $\epsilon > 0$ with $\Ball(y;3\epsilon) \subseteq S$. Put $V=\Ball(y;\epsilon)$. By metric equi-continuity, there exists $U \ni x$ with $f[U] \subseteq \Ball(f(x);\epsilon)$ for all $f\in F$. Assume that $f[U]\cap V\ne\emptyset$. Pick any point $z\in f[U]\cap V $. Since $f[U] \subseteq \Ball(f(x);\epsilon)$, $d(f(x),z)<\epsilon$.  Let $t\in f[U]$ be an arbitrary point. Then 
$$d(t,y)\le d(t,f(x))+d(f(x),z)+d(z,y)<\epsilon+\epsilon+\epsilon=3\epsilon.$$
Thus $f[U]\subseteq \Ball(y;3\epsilon) \subseteq S$.
Topological $\not\Rightarrow$ Metric: Let $X=Y=(0,+\infty)$ endowed with standard topology,  $F=\{f_n:n\in\Bbb N\}$, where $f_n(t)=nt$ for each $t\in (0,+\infty)$, and $x=1$. Clearly, that the family $F$ is not metric equi-continuous at $x$. But it is topologically equi-continuous at $x$. Indeed, let $y\in Y$ be an arbitrary point and $S$ be an arbitrary open neighborhood of $y$. There exists a natural number $N$ such that $f_n(x’)> y+1$ for each $n>N$ and $x’>1/2$. For each $n\le N$ there exists neighborhoods $U_n$ of $1$ and $V_n$ of $y$ such that $nU_n\subseteq V_n\cap S$ provided $y=n$ and $nU_n\cap V_n=\emptyset$, otherwise. Put $U=(1/2,+\infty)\cap\bigcap U_n$ and $V= (0,y+1)\cap S\cap\bigcap V_n$. Assume that $f_n\in F$, $x’\in U$ and $f_n(x’)\in V$. Since $x’>1/2$ and $V\subseteq (0,y+1)$,  $n\le N$. Since $x’\in U_n$ and $nx’\in V_n$, $y=n$. Then $f_n(U)\subseteq f_n(U_n)\subseteq V_n\cap S$.
Wikipedia: “For metric spaces, there are standard topologies and uniform structures derived from the metrics, and then these general definitions are equivalent to the metric-space definitions”. [This claim needs references to reliable sources. (February 2007)] 
:-)
