Let $M$ be a compact ($\mathcal{C}^2$) hypersurface of $\mathbb{R}^k$. Then $M$ is the global level set of a function $f$ having $0$ as a regular value (see this related question).
I think that $f$ can be constructed using the Jordan-Brouwer theorem and the tubular neighborhood theorem. (Define $f$ to be a signed distance to $M$, negative in the bounded connected component of $\mathbb{R}^k \backslash M$.)
But is there a more elementary proof of this?
No, there isn't a more elementary proof. This is because this because this fact and the Jordan-Brouwer theorem can somewhat easily be obtained one from the other. Here is a sketch of the Jordan-Brouwer theorem obtained from $M = f^{-1}(0)$.
Let $M = f^{-1}(0)$ for some smooth function $f : \mathbb{R}^k \rightarrow \mathbb{R}$ having regular value $0$, with $M$ connected. Using the tubular neighborhood theorem, it can be shown that $\mathbb{R}^k$ has at most $2$ connected components. Since $f$ has regular value $0$, the disjoint open sets $B = \{x \,|\, f(x) < 0\}$ and $N = \{x \,|\, f(x) > 0\}$ are non-empty. Thus they are the two connected components of $\mathbb{R}^k \backslash M$. Since $M$ is compact, it can be verified that one of them is bounded.