A $T_{3\frac{1}{2}}$ space is compact if Stone-Weierstrass theorem holds. One proof of this is illustrated in Hewitt's "Certain generalizations of the Weierstrass approximation theorem". A sketch of the proof:
A topological space $X$ is compact iff every proper closed subset is compact. Pick a non-compact proper closed subset $F$ and a point $p$ outside it. Pick an open cover $(U_i)_{i \in I}$ of $F$ which doesn't contain $p$ and has no finite subcover, and refine it so that it contains a local base for every $x \in F$. Also let $(V_j)_{j \in J}$ be a refinement of $\{(F \cup \{p\})^c\}$ such that it contains a local base for every $x \in (F \cup \{p\})^c$. Let $A$ be the subalgebra of $C(X)$ generated by real-valued functions separating $x \in F \cap U_i$ from $U_i^c$ and $x \in V_j$ from $V_j^c$. Then $\overline A$ doesn't contain any real-valued function separating $p$ from $F$, which is a contradiction.
Hewitt mentions that J. C. Oxtoby remarked to him it can be proved by embedding $X$ in $\beta X$. Also in the exercise 3.2.K in Engelking's "General topology", embedding $X$ in $[0, 1]^\kappa$ was given as a hint. I tried picking $p \in \overline{X} \setminus X \subset [0, 1]^\kappa$ but contrary to the proof above, $p$ can't be separated from $X$ since $p \in \overline{X}$. How can one prove the theorem using the hint?
Let $Y$ be any compactification of $X$ and pick a point $p\in Y\setminus X$ and a point $q\in X$. Now consider the subalgebra $A\subset C(Y)\subseteq C_b(X)$ consisting of functions $f$ such that $f(p)=f(q)$. This is a closed unital subalgebra and it separates points of $X$ since $p\not\in X$. However, it is not all of $C_b(X)$, so Stone-Weierstrass fails for $X$.
(Note that in fact if $Y$ is a compactification of $X$ that is different from $\beta X$, then $C(Y)$ itself is a proper subalgebra of $C_b(X)$ and so witnesses that Stone-Weierstrass fails for $X$. However, it is possible that $\beta X$ is the only compactification of $X$ so you need to use the argument above to find a proper subalgebra.)