The Stacks project

Proof. Let $X' \subset X$ be the irreducible component which dominates $Y$. This means that the generic point of $X'$ maps to the generic point of $Y$ (see for example Morphisms, Lemma 29.8.6). By Lemma 37.37.1 we see that $f$ is étale at every point of $X'$. In particular, the open subscheme $U \subset X$ where $f$ is étale contains $X'$. Note that every quasi-compact open of $U$ has finitely many irreducible components, see Descent, Lemma 35.16.3. On the other hand since $Y$ is geometrically unibranch and $U$ is étale over $Y$, the scheme $U$ is geometrically unibranch. In particular, through every point of $U$ there passes at most one irreducible component. A simple topological argument now shows that $X' \subset U$ is both open and closed. Then of course $X'$ is open and closed in $X$ and by connectedness we find $X = U = X'$ as desired. $\square$