Are the nontrivial zeroes of the Riemann zeta function countable?

If the set $Z$ of zeroes of $\zeta(s)$ were uncountable, then it would have an accumulation point. Now, by certain version of identity theorem, this implies that $\zeta(s)$ is identically zero on its domain, which is absurd.

The comment has left me a bit less willing to commit to the statement of my original answer, though the spirit remains, in essence, true since the two are intimately related, but as I know of no other proof of the explicit formula without zero density arguments, I think it best to avoid that as an a priori reason.

The reasons given in the other answers are my other standard go-to explanations, so I'll try and make this accepted one more developed to do justice to the status of "accepted," as the lazy approach (simple deletion) is not available to me.

The problem with uncountably many zeros is that we may cover $\Bbb C\setminus\{1\}$ by a countable number of compact sets, eg. by $A_n, K_n$ which are closed annuli centered at $1$ of inner radius $n$ and outer radius $n+1$ in the cast $A_N$. The $K_n$ are a cover of the disc

$$D=\{z\in\Bbb C | 0<|z-1|<1\}.$$

which can be taken to be

$$K_n =\left\{z\in\Bbb C : {1\over n+1}\le |z-1|\le {1\over n}\right\}.$$

But the set of zeroes, $S_0$ having no limit points would indicate that $S_0\cap K_n$ is finite (it is discrete and compact), hence $S_0$ may be written as a countable union of finite sets

$$S_0=\bigcup_{m}A_m\cap S_0\sqcup\bigcup_n K_n\cap S_0$$

and so is itself finite. This fact is not special about the $\zeta$ function, it holds for any holomorphic function on any relatively nice set, eg. $\Bbb C\setminus\{1\}$.

The original, accepted answer

The set must be countable by a simple argument. If you look at the so-called "explicit formula" it involves a sum over non-trivial zeroes of the $\zeta$ function.

$$\Psi(x)=x-\sum_{\rho}{x^\rho\over\rho}-{\zeta'(0)\over\zeta(0)}-{1\over 2}\log\left(1-{1\over x^2}\right).$$

Here the sum, $\rho$ is taken over non-trivial zeros. But we know that any infinite sum for which more than countably infinitely many terms are non-zero yields a divergent series. Since we know the explicity formula converges, it must be that the number of zeroes is countable.

In short, the number of nontrivial zeroes of $\zeta(s)$ with $\lvert \text{Im}(s) \rvert < T$ is on the order of $T \log T$. So there are countably many nontrivial zeroes.