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Let $K \subset \mathbb{R}^n$ be compact and $(f_n)$ a sequence such that $f_n \colon K \to \mathbb{R}^M$ is continuous for all $n ∈ \mathbb{N}$. Suppose that $(f_n)$ is pointwise convergent and equicontinuous. Show that $(f_n)$ converges uniformly in $K$.

I used the Arzelà–Ascoli Theorem to show that there is a subsequence that converges uniformly, but I didn't know how to extend to the whole sequence, because when I use the upper bound $|f_n(x)-f(x)|<|f_n(x)-f_{n_k}(x)|+|f_n(x)-f(x)|$, the first term in the left-hand-side will need a $\delta$ that depends on $x$.

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$f_n \rightarrow f$ pointwise. You proved that there is a subsequence of $f_n$ which converges uniformly to $f$, right?

Instead of applying your theorem to the $f_n$, apply it to an arbitrary subsequence of $f_n$. You will have proved that every subsequence has a subsequence which converges uniformly to $f$. This makes the sequence converge to $f$ uniformly as well. This is a general fact about topological spaces:

Given a sequence $x_n$, if every subsequence has a subsequence converging to $x$, then the sequence converges to $x$.

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Since the $f_n$'s are continuous on $K$ compact, they're uniformly continuous and being the sequence equicontinuous by hypothesis, no $\delta$ depends on $x$.

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