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I'm studying analysis from Baby Rudin, and his proof (Theorem 7.32) is quite long and split up into four parts. Is there a nicer proof of the theorem?

For reference, a statement of the theorem from the book.

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    I doubt there is a significantly simpler proof. There is no getting around the fact that you have to use the hypotheses to somehow construct a bunch of elements of $\mathscr{B}$, and that's going to take some work. – 2017-02-02
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    Conceptually, in fact, I like to think of Stone-Weierstrass as a lemma that does all this messy work once and for all and lets you prove a bunch of other theorems (e.g., Gelfand duality) without having to get your hands dirty. – 2017-02-02
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    I have once (about 30 years ago) been introduced to a _very_ short proof (http://www.univie.ac.at/nuhag-php/bibtex/open_files/ra84_T%20J%20Ransford.pdf) of the Machado-Bishop-Stone-Weierstrass theorem, which has the Stone-Weierstraß theorem as a co-co-corollary. I was not able to see the idea behind the proof, though. – 2017-02-02
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    The proof in Rudin is actually quite nice. – 2017-02-02

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Indeed the long proof seems to be needed. At least in the framework presented at the book.

An attempt of an alternative shorter proof is presented bellow. The explanation of why the proof is wrong is presented immediately after it.

Rudin's Theorem 7.31, can be easily extended to an arbitrary number of points. Hence, the following result holds:

Theorem 7.31 (Extended). Suppose $\mathcal{A}$ is an algebra of functions on a set $E$, $\mathcal{A}$ separates points on E and vanishes at no point of E. Supose $x_i$, $i = 1, \cdots, N$ are distict points and $c_i$, $i = 1, \cdots, N$, are constants. Then $\mathcal{A}$ contains a function $f$ such that $$f(x_i) = c_i, \text{ for }i = 1, \cdots, N$$

Using this result we could prove step 4 of theorem 7.32 without having to prove step 1, 2 or 3.

Step 4. Given a real function $f$, continuous on K, and $\epsilon>0$, there exists a function $h \in \mathcal{B}$ such that: $$|h(x) - f(x)| < \epsilon (x \in K)$$

Proof: For a set of points $x_i$, $i = 1, \cdots, N$, we can choose h(x) such that $h(x_i) = f(x_i)$, because of the extended theorem 7.31 (above). By the continuity of $f$ and $h$ it follows that there exist a open sets $V_i$, such that: $$|h(t)-f(t)|<\epsilon, \text{ for }t \in \bigcup_{i=1}^N V_i.$$ Since $K$ is compact, there exist a finite set of points $x_i$, $i = 1, \cdots, N$ such that $K \subset \bigcup_{i=1}^N V_i$.

And, as mentioned by Rudin:

Since $\mathcal{B}$ is uniformly closed, this (step 4) is equivalent to the conclusion of the theorem.

This proof, however, is wrong because the set $V_i$ doesn't depend only on $x_i$, but also on all the other $x_j$,$j\not=i$. Therefore you cannot claim, through compactness, that K will be contained in a finite union of those