Let $X$ and $Y$ be schemes locally of finite type over an algebraically closed field $K$, and suppose $Y$ is etale.
If $f:X \to Y$ is a surjective morphism of schemes over $K$, then is the induced map on $K$-points $X(K)\to Y(K)$ also surjective?
My attempt: $Y$ has the discrete topology (etale over an algebraically closed field). The (open) fiber of any $y\in Y$ must contain a $K$-point since the set of $K$-points of $f^{-1}(y)$ is very dense, hence not empty.
Your attempt is basically correct, but could justify a few more things. Here's an explanation which is maybe a little over-justified.
As $X\to\operatorname{Spec} k$ is locally of finite type, $X\to Y$ is locally of finite type (ref), so any base change of $X\to Y$ is again locally of finite type (ref). Since the schemes etale over a field $k$ are $\coprod_{i\in I} \operatorname{Spec} K_i$ for $k\to K_i$ finite separable (ref), we have that $Y=\coprod_{i\in I} \operatorname{Spec} k$ as $k$ is algebraically closed. As surjectivity is preserved under base change (ref), the base change $X\times_Y \operatorname{Spec} k \to \operatorname{Spec} k$ is a surjective morphism locally of finite type, so $X\times_Y \operatorname{Spec} k$ is a nonempty scheme locally of finite type over an algebraically closed field $k$ and therefore it has a $k$-point: on a nonempty affine open, we're looking at the spectrum of a nonzero finitely-generated $k$-algebra, which has a maximal ideal, and by Zariski's lemma the residue field at this closed point is a finite extension of $k$ and therefore $k$ itself. Therefore there is a $k$-point of $X\times_Y \operatorname{Spec} k$, which induces a $k$-point of $X$ mapping down to our chosen $k$-point $\operatorname{Spec} k \subset Y$ by the map $X\times_Y \operatorname{Spec} k \to X$. Thus the map $X(k)\to Y(k)$ is surjective.