A wrong corollary of Kolmogorov's $0-1$ Law

By the Hewitt–Savage 0-1 law, $\limsup S_n$ is a constant $C\in[-\infty,\infty]$ ($\because \limsup S_n$ is measurable w.r.t. the exchangeable $\sigma$-algebra). Since $S_n\overset{d}{=} S_{n+1}-X_1$, $C=C-X_1$ a.s. There are two possibilities for the last equality to hold: (1) $|C|<\infty$ and $X_1=0$ a.s., (2) $C=\pm\infty$. The same is true for $\liminf S_n$.

If $S_n$ converges a.s. then $\limsup S_n = \liminf S_n$ a.s.

So far, so good.

Does this mean that if $S_n$ converges it does so to a constant?

That the answer is "yes" seems surprising, until you remember that a conditional statement "If $P$, then $Q$" is vacuously true when $P$ is false.

In your example, $S_n$ does not converge unless each of the $X_n = 0$ a.s., in which case, $S_n = 0$ a.s. for all $n$ as well and therefore we trivially have $S_n \to 0$ a.s.

If $P(X_n = 0) \neq 1$, then since $S_n$ does not converge, the "if" part of "if $S_n$ converges it does so to a constant" is false, and so the conditional statement is vacuously true.