Every real-valued continuous function on a closed set of compact Hausdorff space has an extension. 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 $\phi(C(X))=C(V)$.
 A: I think this is a special case of the Tietze extension theorem
(since any compact Hausdorff space is normal). (Here is one proof.)
A: If you like to see the Tietze theorem from a Functional Analysis viewpoint, another possibility is by means of the following basic lemma, stating that a certain "approximate surjectivity" actually implies "surjectivity"; this allows to prove a kind of dual result too.

Iteration Lemma. Let $X$ and $Y$ be Banach spaces, $T$ a linear bounded
  operator, $B_X$ the unit closed ball
  of $X$, $U$ a bounded nbd of $0\in Y$, 
  $0\le t <1$ such that $$U\subset
> TB_X+tU.$$ Then, $T$ is onto and in
  fact $$(1-t)U\subset TB_X\\ .$$

You may have nice time applying it to prove the following results, in which we assume wlog that the functions are bounded, so that we can work with the Banach space $C_b$.


*

*Tietze-Dugundji extension theorem. Let $M$ be a metric space, $V\subset M$ closed, $E$ a Banach space. Then the restriction operator $C_b(M,E)\to C_b(V,E)$ is surjective.

*Bartles-Groves lifting theorem. Let $L:E\to F$ a bounded linear surjective operator; $M$ a metric space. Then the composition operator $C_b(M,E)\to C_b(M,F)$, such that $f\mapsto L\circ f,$ is surjective. 
