Is the set of closed points of a $k$-scheme of finite type dense?

Let $k$ be a field.
Let $X$ be a scheme of finite type over $k$.
We denote by $X_0$ the set of closed points of $X$.
Is $X_0$ dense in $X$?

Motivation
See my comment to Martin Brandenburg’s answer to this question.

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Yes, the subset $X_0\subset X$ of closed points is dense in $X$: this is even true if you only assume that $X$ is a $k$-scheme locally of finite type.
Technically the subset $X_0$ has the property of being very dense in $X$, which means that the map sending an open subset of $X$ to its trace on $X_0$ is a bijection between the open subsets of $X$ and those in $X_0$ (provided with the induced topology) .
The proof is not very difficult and uses the characterization of a closed point $x$ as one which has finite dimensional residual field, namely $[\kappa(x):k]\lt \infty$.
A detailed proof can be found in Görtz-Wedhorn’s book, Proposition 3.35.

Edit: warning !
Beginners (and I obviously don’t mean Makoto here!) should be astonished to read that the circle $x^2+y^2+1=0$, seen as a subset of $\mathbb A^2_\mathbb R$, has a dense subset of real closed points!
The paradox is solved by realizing that a closed point of that circle is not an inexistent pair $(r_1,r_2)\in \mathbb R^2$ satisfying $r_1^2+r_2^2+1=0$, but a maximal ideal in $\mathbb R[X,Y]$ containing $X^2+Y^2+1$, like for example the maximal ideal $(X-2,Y^2+5)\subset \mathbb R[X,Y]$.

Yes, this follows from the fact that such a scheme is Jacobson, because a field is Jacobson, and a finite type algebra over a Jacobson ring is Jacobson. One of the characterizations of a Jacobson space is that every closed subset is the closure of its subset of closed points.

Lemma 1
Let $k$ be a field.
Let $A$ be an integral domain which is a $k$-algebra of finite type.
Let $\bar k$ be an algebraic closure of $k$.
Then there exists $k$-algebra homomorphism $\psi\colon A \rightarrow \bar k$.

Proof:
This is a special case of this question(take $b = 1$).

Lemma 2(weak Nullstellensatz)
Let $k$ be a field.
Let $A$ be a $k$-algebra of finite type.
Let $\mathfrak{m}$ be a maximal ideal of $A$.
Then $A/m$ is finite over $k$.

Proof:
Let $\bar k$ be an algebraic closure of $k$.
By Lemma 1, there exists $k$-algebra homomorphism $\psi\colon A/\mathfrak{m} \rightarrow \bar k$.
Since $A/\mathfrak{m}$ is a field, $\psi$ is injective.
Hence $A/\mathfrak{m}$ is algebraic over $k$.
Since it is of finite type over $k$, it is finite over $k$.
QED

Lemma 3
Let $k$ be a field.
Let $X$ be an affine $k$-scheme of finite type.
Let $x \in X$.
Then $k(x)$ is algebraic over $k$ if and only if $x$ is a closed point of $X$.

Proof:
Suppose $X = Spec(A)$, where $A$ is a $k$-algebra of finite type.
Let $P$ be a prime ideal which corresponds to $x$.
We can identify $k(x)$ with the field of fractions of $A/P$.

Suppose $x$ is a closed point of $X$.
Then $P$ is a maximal ideal.
Hence $k(x) = A/P$ is algebraic over $k$ by Lemma 2.

Conversely suppose $k(x)$ is algebraic over $k$.
Then $A/P$ is a finite algebra over $k$.
Since $A/P$ is an integral domain, $A/P$ is a field.
Hence $P$ is a maximal ideal.
QED

Lemma 4
Let $k$ be a field.
Let $X$ be a $k$-scheme locally of finite type.
Let $U, V$ be affine open subsets of $X$ such that $U\cap V \ne \emptyset$.
Let $x \in U \cap V$.
Suppose $x$ is a closed point of $U$.
Then $x$ is also a closed point of $V$.

Proof:
This follows immediately from Lemma 3.

Lemma 5
Let $k$ be a field.
Let $X$ be a $k$-scheme locally of finite type.
Let $U$ be a non-empty affine open subset of $X$.
Then any closed point of $U$ is a closed point of $X$.

Proof:
Let $x$ be a closed point of $U$.
Let $Z$ be the closure of $\{x\}$ in $X$.
Let $\{U_i\}$ be an affine open cover of $X$.
Since $Z = Z \cap X = \bigcup_i Z \cap U_i$, it suffices to prove that $Z \cap U_i = \{x\}$ whenever$Z \cap U_i$ is non-empty.
Suppose $Z \cap U_i$ is non-empty.
Then $x \in U_i$.
By Lemma 4, $x$ is a closed point of $U_i$.
Hence $Z \cap U_i = \{x\}$
QED

Proposition
Let $k$ be a field.
Let $X$ be a $k$-scheme locally of finite type.
Let $X_0$ be the set of closed points of $X$.
Then $X_0$ is dense in $X$.

Proof:
This follows immediately from Lemma 5.