Let $X$ be a compact metric space, $Y$ be a metric space and $\mathcal F \subset C(X,Y)$. I have to show that if $\cal F$ is equicontinuous on $X$, then it's uniformly equicontinuous. The hint given is: argue by contradiction.
Work
Suppose that it's not uniformly equicontinuous. Then there exists $\epsilon > 0$ such that:
$$\forall \delta > 0 , \exists f \in \mathcal F, \exists x,y \in X: d(x,y) < \delta \ \& \ d(f(x),f(y)) \ge \epsilon $$
So:
$$\forall n \in \Bbb N, \exists f_n \in \mathcal F, \exists x_n, y_n \in X: d(x_n,y_n) < 1/n \ \& \ d(f_n(x_n),f_n(y_n)) \ge \epsilon$$
So we may choose sequences $(x_n)$, $(y_n)$ and $(f_n)$ as described above and we have:
$d(x_n,y_n) < 1/n$ for all $n \in \Bbb N$
$d(f_n(x_n),f_n(y_n)) \ge \epsilon$, for all $n \in \Bbb N$ $(\star)$
By sequential compactness, the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$, converging to some $x_0 \in X$. As $d(x_{n_k},y_{n_k}) < 1/n_k$, we get $y_{n_k} \to x_0$ as well.
$\cal F$ is equicontinuous at $x_0$, so there exists $\delta > 0$ such that:
$$\forall f \in \mathcal F, \forall x \in X, d(x,x_0) < \delta \implies d(f(x),f(x_0))< \epsilon/2$$
Since $x_{n_k} \to x_0$ and $y_{n_k} \to x_0$, there is $k_0$ such that:
$$\max \{d(x_{n_{k_0}}, x_0),d(y_{n_{k_0}},x_0)\} < \delta$$
So we get:
$$d(f_{n_{k_0}}(x_{n_{k_0}}), f_{n_{k_0}}(x_0)) < \epsilon /2$$
and
$$d(f_{n_{k_0}}(y_{n_{k_0}}), f_{n_{k_0}}(x_0)) < \epsilon /2$$
Then:
$$d(f_{n_{k_0}}(x_{n_{k_0}}), f_{n_{k_0}}(y_{n_{k_0}}) < \epsilon/2 + \epsilon/2 = \epsilon$$
which is a contradiction to $(\star)$.
What do you think about this proof?