Geometric interpretation of regular non-closed points Let $X$ be a variety (say, integral scheme of finite type) over an algebraically closed field $k$. Consider an irreducible closed subset $Y\subseteq X$ and let $\eta$ be the generic point of $Y$. I would like to know if the following is true.

Question: The point $\eta$ is regular (i.e, $\mathcal{O}_{X,\eta}$ is a regular local ring) if and only if there is at least one closed point $x\in Y$ that is regular on $X$.

The if part of the question is true due to a theorem of Serre: The localization of a regular local ring is a regular local ring. Because, if the variety $Y$ is defined by the prime ideal $P$ inside an affine chart containing $x$, then $(\mathcal{O}_{X,x})_{P\mathcal{O}_{X,x}}=\mathcal{O}_{X,\eta}$.
So I am interested in the only if part. 
Notice that any algebraic variety has at least one regular closed point (cf. the book of Liu, section 4.2 Lemma 2.21, where it is proved over any field under the assumption of $X$ being geometrically reduced). Hence, $Y$ has at least one closed point $x$ such that $\mathcal{O}_{Y,x}$ is regular. As we are assuming that $\mathcal{O}_{X,\eta}$ is regular as well maybe some kind of transitivity of the regular property (in the sense of this other question that I asked before) would imply that the point $x$ is regular in $X$. Maybe the fact that the set of regular points in $Y$ is actually an open set (cf. Liu section 4.2 proposition 2.24) can be used to change the point $x$ in case some of them fails.
If you find a nice proof of this it would be nice to clean the unnecessary hypothesis as well.
 A: I believe what is going on is point-set topology, because of the following:
Lemma. Let $X$ be a Jacobson space and consider a locally quasi-constructible subset $U \subseteq X$. If $Y \subseteq X$ is a locally quasi-constructible subspace such that $U \cap Y \ne \emptyset$, then there exists a point $x \in Y$ that is closed in $X$ such that $x \in U$.
Proof. The set $U \cap Y$ is locally quasi-constructible in $Y$ by definition of the subspace topology, and is non-empty by hypothesis. Since $Y$ is a Jacobson topological space [EGAI$_{\text{new}}$, Chapitre 0, Proposition 2.8.2], we see that $U \cap Y$ contains a closed point $x \in Y$ by definition of a Jacobson space [EGAI$_{\text{new}}$, Chapitre 0, Définition 2.8.1]. This point $x$ is closed in $X$ by [EGAI$_{\text{new}}$, Chapitre 0, Proposition 2.8.2]. $\blacksquare$
To apply this to your situation, every scheme of finite type over a field $k$ is Jacobson [EGAI$_{\text{new}}$, Proposition 6.5.2], and setting $U$ to be the regular locus $\operatorname{Reg}(X)$, we see that $\operatorname{Reg}(X)$ is open, hence locally quasi-constructible.
Of course, maybe it is easier to just prove your statement directly: The set $\operatorname{Reg}(X) \cap Y$ is open in $Y$ by definition of the subspace topology, and is non-empty since it contains $\eta$. Since closed points are dense in $Y$ [EGAI$_{\text{new}}$, Proposition 6.5.2], there exists a point $x \in \operatorname{Reg}(X) \cap Y$ that is closed in $Y$. Finally, this point $x$ is closed in $X$ since $\{x\} \hookrightarrow Y \hookrightarrow X$ is a composition of closed embeddings.
A: For the sake of future reference let me spell out a version of Takumi's result and indicate why it may fail in a different context.
Takumi's result
Let $X$ be a scheme locally of finite type over a field $k$ or over $\mathbb Z$, let $\operatorname {Reg }(X)\subset X$ be the set of regular points and let $\operatorname {Reg }^{cl}(X)\subset \operatorname {Reg }(X)$ the set of closed regular points.
Then for any point $\eta\in X$ we have the equivalence: 
$$\boxed {\mathcal O_{X,\eta} \operatorname {is a regular local ring} \iff
\overline {\{\eta\}}\cap  \operatorname {Reg }^{cl}(X)\neq \emptyset  }$$ However the result is no longer true for arbitrary schemes:
A very simple counter-example
Let $k[x,y]=\frac {k[X,Y]}{\langle X\cdot Y\rangle}, \mathfrak m={\langle x, y\rangle}$ and $A=k[x,y]_\mathfrak m$, a singular local ring of dimension one.
Our counter-example concerns the scheme $X=\operatorname {Spec}A=\{ M,\eta, \xi\}$, where $M$ is the closed point corresponding to $\mathfrak m$ and  $\eta =\langle  y\rangle, \xi=\langle  x\rangle$ are the two open points of $X$.
We have $\operatorname {Reg }^{cl}(X)=\emptyset$, and so a fortiori $\overline {\{\eta\}}\cap \operatorname {Reg }^{cl}(X)= \emptyset  $.
However $\mathcal O_{X,\eta}$ is a regular local ring since it is the rational function field $k(x)$.
This counter-example is only possible because $A_\mathfrak m$ is not of finite type over $k$, and Takumi's assumption is thus not satisfied.
An even simpler counter-example
Let $k[x,y]=\frac {k[X,Y]}{\langle Y^2-X^3\rangle}, \mathfrak m={\langle x, y\rangle}$ and $A=k[x,y]_\mathfrak m$, a singular local ring of dimension one.
Our counter-example concerns the scheme $X=\operatorname {Spec}A=\{ M,\eta\}$, where $M$ is the closed point corresponding to $\mathfrak m$ and  $\eta =\langle  0\rangle$ is the  generic point of $X$, which is open.
We have $\operatorname {Reg }^{cl}(X)=\emptyset$, since the only closed point $M$ of $X$ is singular.
So a fortiori $\overline {\{\eta\}}\cap \operatorname {Reg }^{cl}(X)= \emptyset  $.
However $\mathcal O_{X,\eta}$ is a regular local ring since it is the field $k(x,y)=\operatorname {Frac}A=\operatorname {Frac}k[x,y]$.
Hence the boxed equivalence fails, and the reason again is that $X$ is not locally of finite type over $k$.
