Let $k$ be a field, and let $R$ be an integral domain and finitely generated $k$-algebra. It is a basic result from commutative algebra that the dimension of $R$ is equal to the transcendence degree of the quotient field of $R$ over $k$.
In particular, if $X = \textrm{Spec } A$ is an affine scheme of finite type over $k$, and $\mathfrak p \in X$, then the dimension of $A/\mathfrak p$ is equal to the transcendence degree of the residue field $\kappa(x) = A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$ over $k$. In particular, the point $\{\mathfrak p\}$ is closed if and only if $\mathfrak p$ is a maximal ideal of $A$, if and only if $\textrm{tr.deg}_k\kappa(x)=\textrm{Dim } A/\mathfrak p = 0$, if and only if $\kappa(x)$ is algebraic over $k$.
Now, assume that $X$ is a scheme which is locally of finite type over $k$. This means that $X$ has an open cover $U_i : i \in I$ by affine open sets, such that each $\mathcal O_X(U_i)$ is finitely generated as a $k$-algebra. In fact, every affine open subset of $X$ is the spectrum of a ring which is finitely generated as a $k$-algebra.
Proposition: The following are equivalent for $x \in X$:
(i): $x$ is a closed point.
(ii): $x$ is a closed point in some affine open set containing it.
(iii): $x$ is closed in every affine open set containing it.
(iv): $\kappa(x)$ is algebraic over $k$.