I just came up with the following proof. What I like about it is that when I came up with it I felt like I was just following my nose. It's somewhat different from the other ones I've seen, so I'd like to know if it's correct, and whether someone has seen it before.
Let $f:X\to Y$ be a continuous function where $X, Y$ are metric spaces with metrics $d_X$ and $d_Y$ respectively, and $X$ is compact. I want to prove that $f$ is uniformly continuous.
From continuity we have that for all $x_0\in X$ and all $\epsilon\in\mathbb R_{>0}$, there exists a $\delta>0$ so that
$$d_X(x,x_0)<\delta\implies d_Y(f(x), f(x_0))\leq\epsilon.$$
So for each $\epsilon$ define a function $\delta_\epsilon:X\to \mathbb R_{>0}$ by
$$\delta_\epsilon(x_0)=\sup_{\delta\in\mathbb R_{>0}}\left(\delta: d_X(x,x_0)<\delta\implies d_Y(f(x), f(x_0))\leq\epsilon\right).$$
$\delta_\epsilon$ is continuous† and its domain is compact so there is an $m\in X$ so that $\delta_\epsilon(x)\geq \delta_\epsilon(m)\forall x\in X$. Then $\delta_\epsilon(m)$ is the delta we need for the definition of uniform continuity.
† Suppose $\delta_\epsilon$ is not continuous at $x_0\in X$, then there's a sequence $x_1, x_2\dots\to x_0$ so that $\delta_\epsilon(x_n)\to l\neq \delta_\epsilon(x_0)$. Suppose $l>\delta_\epsilon(x_0)$, and the other case is analogous. Choose $n$ big enough so that $\delta_\epsilon(x_n)>\delta_\epsilon(x_0)$ and $d_X(x_n, x_0)<\frac{\delta_\epsilon(x_n)-\delta_\epsilon(x_0)}2$ which means that
$$d_X(x,x_0)<\frac{\delta_\epsilon(x_n)+\delta_\epsilon(x_0)}2\implies d_X(x, x_n)<\delta_\epsilon(x_n)$$
$$\implies d_Y(f(x), f(x_n))\leq\epsilon$$
$$\implies d_Y(f(x), f(x_0))\leq d_Y(f(x), f(x_n))+d_Y(f(x_0), f(x_n))\leq\epsilon+d_Y(f(x_0), f(x_n))$$
and by the continuity of $f$ for any $t>\epsilon$ we can choose $n$ sufficiently large so that $d_Y(f(x_0), f(x_n))<\frac{t-\epsilon}2$, which means
$$d_Y(f(x), f(x_0))<\frac{t+\epsilon}2<t$$
$$d_Y(f(x), f(x_0))\leq \epsilon.$$
This holds for all $x$ so that $d_X(x,x_0)<\frac{\delta_\epsilon(x_n)+\delta_\epsilon(x_0)}2$ which contradicts the definition of $\delta_\epsilon(x_0)$.