I've noted, that the following fact can be proven in a few lines using $C^*$-algebra theory. I wonder if it has a simple elementary proof or not. Probably you can give me a reference.
Suppose $X$ is a compact Hausdorff space, $V\subset X$ is a closed subset, $f\colon V\to \mathbb{R}$ is a continuous function. Then there exists a continuous function $g\colon X\to \mathbb{R}$, whose restriction on $V$ is $f$.
C*-algebraic proof is the following.
First note, that it is enough to find $g\colon X\to \mathbb{C}$ (then we can take its real part). Consider restriction map $\phi\colon C(X)\to C(V)$. It is a * -homomorphism, and therefore its image $\phi(C(X))$ is a commutative $C^*$-algebra. Moreover $\phi(C(X))$ separates points of $V$ (because $C(X)$ does). By Stone-Weierstrass theorem $\phi(C(X))=C(V)$.
Best Answer
I think this is a special case of the Tietze extension theorem (since any compact Hausdorff space is normal). (Here is one proof.)