How can one prove that manifolds are regular?

Note that locally Euclidean and Hausdorff implies locally compact Hausdorff. So the following more general result will answer your question.

Every locally compact Hausdorff space is completely regular. A proof can be found here.

The main idea is that the locally compact Hausdorff spaces are precisely the spaces which admit a one-point (or "Alexandroff") Hausdorff compactification. Now compact Hausdorff spaces are normal, hence completely regular. Normality need not be inherited by an arbitrary subspace, but complete regularity is.

[Note: in general I am a fan of the convention that "compact" and "locally compact" include the Hausdorff condition. So as to be maximally transparent, I am -- clearly -- not imposing that convention here.]

Pete's answer is already sufficient for anyone who wants to work out the details of the essential lemma that locally Euclidean and Hausdorff implies locally compact Hausdorff on their own, so I wanted to add in a proof of that lemma for completeness of this page.

Proof: Let $M$ be locally Euclidean. We claim that for each point $x\in M$ there is a compact set $K\subset X$ and a neighborhood $U\subset K$ of $x$. Fix $x\in M$; since $M$ is locally Euclidean, there is a neighborhood $N$ of $x$, an open set $V\subset \mathbb R^n$, and a homeomorphism $f:N\to V$. Choose some ball $B\subset V$ centered at $f(x)$ whose closure $\overline B$ lies entirely inside of $V$, and consider the images $f^{-1}(B),f^{-1}(\overline B)$ of these two sets under the inverse homeomorphism $f^{-1}:V\to N$. Since $f^{-1}$ is an open map, $f^{-1}(B)$ is a neighborhood of $x$; since $f^{-1}$ is continuous, $f^{-1}(\overline B)$ is compact in $M$ and $f^{-1}(B)\subset f^{-1}(\overline B)$. This proves the claim.