closed point of a locally finite $k$-scheme Let $X$ be a locally finite $k$-scheme, where $k$ is a field. 
Suppose I have $Spec B \subseteq X$ such that $B$ is a finitely generated
$k$-algebra, and $p \in Spec B$ a closed point inside $Spec B$ with respect to
the subspace topology. Does it then follow that $p$ is in fact a closed point of X?
Thanks! 
 A: Yes, that's true. If $X$ is locally of finite type over $k$, then $x\in X$ is closed in $X$ if and only if $k(x)/k$ is a finite extension.
"$\Rightarrow$": We may assume $X=\text{Spec}(A)$ affine, so $x={\mathfrak m}$ is a maximal ideal. Then $A/{\mathfrak m}$ is a field extension of finite type, hence finite. 
"$\Leftarrow$": Since $X$ is covered by affine open subsets of the form $\text{Spec}(A)$ with $A$ of finite type over $k$, and since being closed can be checked on open coverings, we may assume that $X=\text{Spec}(A)$. Then $x={\mathfrak p}$ is a prime in $A$ such that $\text{Quot}(A/{\mathfrak p})/k$ is finite. In particular, $A/{\mathfrak p}$ is algebraic over $k$, and hence $A/{\mathfrak p}$ is a field.
As a corollary we obtain that $x\in X$ is closed in $X$ if and only if it is closed in some neighbourhood.
Without $X$ being locally of finite type over a field, the statement is false. For example, consider the closed point $(\pi T - 1)$ in $\text{Spec}(R[T])={\mathbb A}_R^1$ for $R$ a discrete valuation ring with prime $\pi$: Its image under ${\mathbb A}_R^1\to\text{Spec}(R)$ is the (non-closed) generic point $(0)$. Hence, if $x$ is considered as a point in ${\mathbb P}^1_R$ through the open embedding ${\mathbb A}^1_R\subset{\mathbb P}^1_R$, it is locally closed but not closed, since the projection ${\mathbb P}^1_R\to\text{Spec}(R)$ is closed.
