[proofplan] The forward direction covers a compact set by finitely many discs of uniform convergence (using compactness) and takes the worst convergence rate among them. The reverse direction uses the fact that every closed disc in $U$ is compact, so uniform convergence on compacta immediately gives local uniform convergence. [/proofplan] [step:Cover the compact set by finitely many convergence discs ($\Rightarrow$)] Suppose $f_n \to f$ locally uniformly on $U$. Let $K \subset U$ be compact. For each $a \in K$, the locally uniform convergence hypothesis provides $r_a > 0$ with $\overline{B(a, r_a)} \subset U$ and $f_n \to f$ uniformly on $B(a, r_a)$. The collection $\{B(a, r_a) : a \in K\}$ is an open cover of $K$. Since $K$ is compact, extract a finite subcover $B(a_1, r_1), \ldots, B(a_m, r_m)$. Given $\varepsilon > 0$, for each $j \in \{1, \ldots, m\}$ there exists $N_j \in \mathbb{N}$ such that $|f_n(z) - f(z)| < \varepsilon$ for all $z \in B(a_j, r_j)$ and all $n \geq N_j$. Set $N = \max\{N_1, \ldots, N_m\}$. For $n \geq N$ and any $z \in K$, the point $z$ belongs to some $B(a_j, r_j)$ in the finite subcover, so $|f_n(z) - f(z)| < \varepsilon$. Hence $f_n \to f$ uniformly on $K$. [guided] The key point is that locally uniform convergence gives a disc of uniform convergence around each point, but the rate of convergence (the index $N$) may differ from disc to disc. For a compact set, we only need finitely many discs, so we can take the maximum of finitely many indices. For each $a \in K$, choose $r_a > 0$ with $f_n \to f$ uniformly on $B(a, r_a) \subset U$. The open balls $\{B(a, r_a)\}_{a \in K}$ cover $K$. Compactness of $K$ extracts a finite subcover $B(a_1, r_1), \ldots, B(a_m, r_m)$. Given $\varepsilon > 0$, each $B(a_j, r_j)$ gives an index $N_j$ with $\sup_{z \in B(a_j, r_j)}|f_n(z) - f(z)| < \varepsilon$ for $n \geq N_j$. Setting $N = \max_j N_j$: for $n \geq N$ and $z \in K \subset \bigcup_j B(a_j, r_j)$, we have $|f_n(z) - f(z)| < \varepsilon$. Why does this argument fail for non-compact $K$? The cover would require infinitely many discs, and the supremum of infinitely many $N_j$ might be infinite --- there would be no single $N$ that works for all of $K$. [/guided] [/step] [step:Use compactness of closed discs for the reverse direction ($\Leftarrow$)] Suppose $f_n \to f$ uniformly on every compact subset of $U$. Fix $a \in U$ and choose $r > 0$ with $\overline{B(a, r)} \subset U$. The closed disc $\overline{B(a, r)}$ is a compact subset of $U$, so $f_n \to f$ uniformly on $\overline{B(a, r)}$. Since $B(a, r) \subset \overline{B(a, r)}$, the convergence is uniform on $B(a, r)$. As $a$ was arbitrary, $f_n \to f$ locally uniformly on $U$. [/step]