You could give a different proof of compactness as follows: Somehow(!) prove soundness and completeness for some formal proof system: A set is satisfiable if and only if it is consistent. It's clear that $\sum$ is consistent if and only if every finite subset of $\sum$ is consistent: If $\sum\vdash\phi\land\lnot\phi$ then the proof of the contradiction uses only finitely many formulas in $\sum$ (since the proof has only finitely many steps), so there exists a finite set $F\subset\sum$ with $F\vdash\phi\land\lnot\phi$.
Of course that's harder than the proof you're asking about. One can use the Tychonoff theorem from topology to give a proof that's not harder, although it may be harder to see where the proof comes from:
First notation: If $v$ is a truth assignment, in other words a map from propositional variables to $B=\{0,1\}$, let $\overline v$ be the extension of $v$ to all formulas (so $\overline v(\phi\land\psi)=\overline v(\phi)\land\overline v(\psi)$, etc.). (I don't recall what notation Enderton uses for $\overline v$, although I suspect it's just $\overline v$.)
Let $L$ be the set of all wffs, and let $X=B^L$, with the product topology arising from the discrete topology on $B$. The Tychonoff theorem shows that $X$ is compact. Note that an element of $X$ is precisely a function $f:L\to B$.
Let $V$ be the set of $f\in X$ such that $f=\overline v$ for some assignment $v$. Show that given $f\in X$ we have $f\in V$ if and only if $f(\phi\land\psi)=f(\phi)\land f(\psi)$ for all $\phi$, $\psi$, and similarly for the other logical connectives. Hence $V$ is a closed subset of $X$, so $V$ is compact.
Now for $E\subset L$ let $S_L$ be the set of $f\in V$ such that $f(\phi)=1$ for all $\phi\in E$. Show that $S_E$ is closed, hence compact.
Say $\sum\subset L$ is finitely satisfiable. We need to show that $\sum$ is satisfiable, which is to say $S_\sum\ne\emptyset$. We're given that $S_F\ne\emptyset$ for every finite set $F\subset \sum$. Since $$\bigcap_{j=1}^nS_{F_j}=S_{\bigcup_{j=1}^nF_j}$$it follows that the intersection of finitely many $S_F$ is nonempty.
So compactness shows that the intersection of all the $S_F$, for $F\subset \sum$ finite, is nonempty. So $$S_\sum=\bigcap_{F\subset\sum}S_F\ne\emptyset,$$qed.
So yes, there are other proofs. I see that while I was typing this someone else posted an answer talking about maximal consistent sets, so I'll stop here.