Are there $k$-rational points that are not closed point? I know for a scheme $X$ locally of finite type over a field $k$, $k$-rational points are closed ponits. 
If we remove the assumption that $X$ is locally of finite type over $k$, are there some $k$-rational points which are not closed point?
 A: Reducing to the affine case, your question comes down to the following: if $A$ is an algebra over a field $k$, not necessarily finitely generated, and $\mathfrak p$ is a prime of $A$ such that $k \rightarrow A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$ is an isomorphism, is it possible for $\mathfrak p$ to be a nonmaximal ideal?  The answer is no.
By hypothesis, the composition $k \xrightarrow{i} A \xrightarrow{\pi} A/\mathfrak p \xrightarrow{h} \operatorname{Quot}(A/\mathfrak p) = A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$ is an isomorphism.  Since
$$h \circ (\pi \circ i)$$
is a bijection, it is in particular surjective, which implies that $h$ must be surjective.  The surjectivity of
$$A/\mathfrak p \rightarrow \operatorname{Quot}(A/\mathfrak p)$$
implies that $A/\mathfrak p$ is a field, i.e. $\mathfrak p$ is maximal.
A: Let $x\in X$ be a $k$-rational point and $X=\bigcup U_i$ be an affine open cover (not necessarily finite).
Then,
$$
\overline{\{ x\}}^{(X)}=\overline{\{ x\}}^{(X)}\cap \left( \bigcup U_i\right) = \bigcup \left( \overline{\{ x\}}^{(X)}\cap U_i\right)=\bigcup_{U_i\ni x} \left( \overline{\{ x\}}^{(U_i)}\right).
$$
If the affine case is verified, $\overline{\{ x\}}^{(U_i)}=\{x\}$ since $x $ is also $k$-rational in $U_i$. So, 
$$
\overline{\{ x\}}^{(X)}=\{x\}.
$$
Therefore we can reduce to the affine case.
