If $X$ is an arbitrary scheme, I can prove that the set $X(k)$ of $k$-valued points is in bijective correspondence with the set $$\{(x,\iota) \, | \, x \in X, \, \iota:\kappa(x) \hookrightarrow k\},$$ where $\kappa(x)$ is the residue field of $x$ on $X$. (This is the content of Hartshorne, Exercise II.2.7.) Assuming now that $X$ is (locally?) of finite type over $k$, $\kappa(x)$ becomes a finitely generated $k$-algebra, which is a field, hence (by Zariski's Lemma) is a finite extension of $k$.
This seems to imply that, in this case, $X(k)$ bijects with the set $\{x \in X \, | \, \iota:\kappa(x) \to k \mbox{ is an isomorphism}\}$ (such points are $\textit{$k$-rational}$), but $X(k)$ is supposed to be in bijective correspondence with the set of closed points, and the set of closed points merely have the property that $\kappa(x)/k$ is a finite extension. How do I reconcile this? In other words, is it true that for $X$ finite type over $k$ algebraically closed, $k$-rational points coincide with closed points? Certainly, $k$-rational points are closed, but if this is true, how do I show the converse?
If $k$ is not algebraically closed, I have seen examples of closed points which are not $k$-rational. Where have I used the fact that $k$ is algebraically closed?
Thanks so much in advance.