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 $\phi(C(X))=C(V)$.



Every real-valued continuous function on a closed set of compact Hausdorff space has an extension.
I think this is a special case of the [Tietze extension theorem](https://en.wikipedia.org/wiki/Tietze_extension_theorem)
(since any compact Hausdorff space is normal). (Here is [one proof](https://planetmath.org/proofoftietzeextensiontheorem).)


