I just found the following false proof of the (correct!) Skolem-Mahler-Lech theorem, which I think is interesting.

Statement of (correct) theorem: Suppose $f(z) := \sum_{n=0}^{+\infty} a_n z^n \in \mathbb{C}[[z]]$ is rational. Let $b_n$ equal $1$ when $a_n\neq 0$ and $0$ when $a_n = 0$. Then $g(z) := \sum_{n=0}^{+\infty} b_n z^n$ is also rational.

False proof: Since $f$ is defined by a linear recurrence relation, correcting for the uninteresting constant term, we can interpret it as the series recognized by a weighted finite automaton on the unary language (i.e., consisting of words over the single letter $z$; so the automaton is just a digraph with complex “multiplicities” associated to edges, and $a_n$ is the number of paths of length $n$, each counted with a multiplicity given by the product of the multiplicities of the edges, from an initial node to a final node: see, e.g., Bousquet-Mélou, “Rational and algebraic series in combinatorial enumeration”, §2). Now make this automaton deterministic (or at least unambiguous) while forgetting multiplicities: in the new automaton, the number of paths of length $n$ from an initial node to a final node is simply $b_n$, i.e., $1$ or $0$ according as there is or isn't such a path in the original automaton. But for the same reason (backwards), $g$ is now given by a linear recurrence relation, so it is rational.

Comment: The error is simply that when forgetting multiplicities we also forget possible cancellations between them: two paths could have multiplicities summing to zero. But the proof does work, and generalize to more variables, when $f$ is $\mathbb{N}$-rational, because no cancellation is possible: see the correct statements in Berstel & Reutenauer, Noncommutative Rational Series with Applications, esp. chapter 3 lemma 1.4. So the idea of the proof isn't stupid and gives related theorems, and the conclusion as stated is correct, yet the proof probably can't be fixed to yield that exact conclusion (because it would then work over any field, which isn't true), so I think this qualifies as an interesting proof.