Why can the meromorphic function only have finitely many poles in the complex plane?

There is a proof in Complex Analysis by Stein-Shakarchi that I do not understand. I have highlighted it in red.

Why can they say that $f$ only has a finite number of points? What makes them able to exclude the countable number in the definition of a meromorphic function?

Solution 1:

If $f\left(\frac{1}{z}\right)$ is holomorphic in $0 < \lvert z\rvert < r$, then $f(z)$ is holomorphic in $\frac{1}{r} < \lvert z\rvert < \infty$ (and vice versa). Thus $f$ can have poles - except maybe at $\infty$ - only in the compact set $\overline{D_{1/r}(0)} = \{z : \lvert z\rvert \leqslant 1/r\}$. Since the set of poles has no accumulation point, and every infinite subset of a compact set has an accumulation point, $f$ can have only finitely many poles in $\overline{D_{1/r}(0)}$, and possibly one at $\infty$, so finitely many altogether.