This is an example I learned only recently from one of my professors (I don't know much about probability, so I'm not sure whether this is an old hat):
We can use Stone-Weierstrass to prove the Kolmogorov extension theorem (or a version thereof). Namely, given a collection $X = \{\mu_\alpha\}_{\alpha}$ of probability measures on $[0,1]$, there exists a probability measure $\mu$ on $\prod_{\alpha} [0,1]$ such that $\mu(\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) ) = \prod_{i=1}^n \mu_{\alpha_i}(A_i)$ for all $\alpha_1, \dots, \alpha_n$, where $\pi_\alpha: X \to [0,1]$ denotes projection.
Sketch of proof: The set $Q \subset C(X)$ of continuous maps, which depend only on "finitely many components", i.e. those $f\in C(X)$ which can be written in the form $f(x) = g(\pi_{\alpha_1}(x), \dots, \pi_{\alpha_n}(x)) \qquad \text{ for some }\alpha_1, \dots, \alpha_n,\, n\in \mathbb N \text{ and } g\in C([0,1]^n)$ is a unital algebra which separates points. By Tychonoff's theorem we know that $X$ is compact, so $Q$ is dense by Stone-Weierstrass.
We can now define a continuous linear functional $I: C(X) \to \mathbb R$ as follows: Given $f(x) = g(\pi_{\alpha_1}(x), \dots, \pi_{\alpha_n}(x)) \in Q$ as above, we define $I(f) = \int_{[0,1]^n} g(x_1, \dots, x_n) \, dx_1 \dots dx_n$ Then $I$ is well-defined and is a bounded linear functional on the dense subspace $Q\subset C(X)$. Therefore it can be extended uniquely to a bounded linear functional on all of $C(X)$.
Finally, Riesz' representation theorem for compact Hausdorff spaces shows that to every such functional $I$ there corresponds a unique Radon measure $\mu$ such that $I(f) = \int_X f\, d\mu$.
It is now not hard to show that this $\mu$ satisfies $\mu(\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) ) = \prod_{i=1}^n \mu_{\alpha_i}(A_i)$ by finding a suitable sequence of continuous functions approximating the indicator function of $\pi_{\alpha_1}^{-1}(A_1) \cap \dots\cap \pi_{\alpha_n}^{-1}(A_n) $. $\square$
I had seen other proofs of this extension theorem (which I didn't like too much, because I couldn't easily see what's going on), so I was really amazed when I first heard of the above argument. I especially like how various big theorems suddenly pop up and fit together so well! =)