Let $k$ be an algebraically closed field. Let $\phi:\textrm{Spec}\ k\longrightarrow X$ be a morphism of schemes of finite type. Let us denote the only point of $\textrm{Spec }k$ by $\zeta$, and the image of $\zeta$ under $\phi$ be $x$, i.e., $\phi(\zeta)=x$. I want to prove that $x$ is a closed point in $X$.
I know that $x$ is closed point iff $k\subset k(x)$ is a finite extension, where $k(x)$ is the residue field at $x$. So by Zariski's lemma, it is enough to prove that $k(x)$ is a finitely generated $k$-algebra. But how do I prove this? I am guessing that this is a consequence of $\phi$ being a morphism of finite type. But I am not able to get a proof. Any help will be appreciated!