Let $\{f_n\}$ be a equicontinuous sequence and converge pointwise in a compact set $K$ of $\mathbb{R}^n$. Prove that the sequence converge uniformly in $K$.
My attempt: Since $\{f_n\}$ is equicontinuous, given $\epsilon > 0$, there exist a $\delta >0$ such that $$|f_{i}(x)-f_{i}(y)|<\epsilon$$ if $1\leq i \leq n $ and $|x-y|<\delta.$ In fact, we know that, given $\epsilon>0$ there exist $N \in \mathbb{N}$ such that, as $n \geq N$ then $|f_n - f|<\epsilon$. I must prove that the convergence is uniform.
Since continuous functions are uniformly continuous on compact set, then, given $\epsilon>0$ there exist $\delta (\epsilon)>0$ such that, as $|x-y|<\delta$ we have $|f_{i}(x)-f_{i}(y)|<\epsilon$. So, taking $n \geq N, m\geq N$ there exist a $\delta>0$ such that, as $|x-y|<\delta$ then $$|f_{n}(x) - f_{m}(x)|<|f_{n}(x)-f_{n}(y)| + |f_{n}(y)-f_{m}(y)| + |f_{m}(y)-f_{m}(x)|<3\epsilon,$$
where $|f_{n}(y)-f_{m}(y)|<\epsilon$ by pointwise convergence.
Thus, $\{f_n\}$ is Cauchy, and then $\{f_n\} \rightarrow f$ uniformly.
Is this correct?