# Finite étale cover of $\mathbb{P}^1$ has finitely many connected components?

I am reading Hartshorne's proof of $\mathbb{P}^1$ being simply connected as a scheme. It seems one ingredient of the proof is that if $X\rightarrow\mathbb{P}^1$ is an étale covering, then X has only finitely many connected components. But I do not see why.

The important point is that $f:X\to \mathbb{P}^1$ must be a finite map. Finite maps are quasi-finite, so the preimage of any point in $\mathbb{P}^1$ is a finite set. Since the number of points in the preimage is an upper semicontinuous function, there will be a generic multiplicity $n_0$ and finitely many points $p_1,\cdots,p_m$ in $\mathbb{P}^1$ which will have a larger number of points $n_1,\cdots,n_m$ in the fiber. Take the max of the $n_i$, and there are at most that many connected components.
Since $f$ is finite, in particular of finite type, $X$ is a Noetherian scheme. Noetherian schemes are locally connected (see http://stacks.math.columbia.edu/tag/04MF), so the connected components of $X$ are open. They therefore form a covering of $X$ by pairwise disjoint open sets. Because $X$ is quasi-compact, this covering must be finite, i.e., $X$ must have finitely many connected components.