Suppose $\{f_i\}_{i=1}^{\infty}$ is continuous "at once" meaning for every $\epsilon >0$ there is $\delta>0$ such that for every $i \in \mathbb{N}$ and for every $|x-y| < \delta$ we get that $|f_i(x)- f_i(y) |<\epsilon$; $X$ is a compact space for which $f_i : X \to X$ ; and $f_n \to f$ meaning $f(x) = \lim \limits_{n \to \infty} f_n(x)$.
Prove that $f_n \Rightarrow f $, meaning $f_n$ converges uniformly to $f$ ??
Here $X$ is a compact normed space and the sequence $(f_i)$ is uniformly equicontinuous and pointwise convergent.
For all $\epsilon > 0$ there exists $\delta > 0$ such that if $\|x-y\| < \delta$, then for all $i \in\mathbb{N}$ we have $\|f_i(x) - f_i(y)\| < \epsilon/3$. Taking the limit as $i \to \infty$ it follows that $\|f(x) - f(y)\| \leqslant \epsilon/3$ as well.
Since $X$ is compact there exist finitely many points $x_1, \ldots x_n$ such that $X \subset \cup_jB(x_j,\delta),$ where $B(x_j,\delta)$ is the open ball with center $x_j$ and radius $\delta$.
By pointwise convergence, there exists $N_j \in \mathbb{N}$ such that if $i \geqslant N_j$ then $\|f_i(x_j) - f(x_j)\|< \epsilon/3.$ Let $N = \max_{1 \leqslant j \leqslant n}N_j.$
Given any $x \in X$ there exists $j$ such that $x \in B(x_j,\delta)$. This implies $\|x - x_j\| < \delta$ and if $i \geqslant N$ we have
$$\|f_i(x) - f(x)\| \leqslant \|f_i(x) - f_i(x_j)\|+\|f_i(x_j) - f(x_j)\|+\|f(x_j) - f(x)\| \\\leqslant \frac{\epsilon}{3} + \frac{\epsilon}{3}+ \frac{\epsilon}{3} = \epsilon,$$
proving uniform convergence.