Suppose $X$ is a scheme. We say a point $x$ is closed in $X$ if $\overline{\{x\}}=\{x\}$. Let $t(X)$ be the subspace of all closed points in $X$. We say X is irreducible if its topological space is irreducible.
My question is if it is true that $X$ is irreducible if and only if $t(X)$ is irreducible.
And in general if $X$ is only a topological space, $t(X)$ is subspace of all closed points in $X$, if it is true that $ X$ is irreducible if and only if $t(X)$ is irreducible
I try to prove it is right. But I spend a lot of time and do not know how to prove it.
This is not true!
$t(X)$ irreducible does not imply $X$ irreducible: Let $ X = \operatorname{Spec} k[x,y]/(xy)_{(x,y)}$ (taking the stalk of the union of two lines at the origin). Your points here are prime ideals of $k[x,y]$ containing $(xy)$ contained in $(x,y)$. The only such prime ideals are $(x,y), (x)$ and $(y)$. You have a unique maximal ideal $(x,y)$, so $t(X)$ is irreducible. But $X$ is reducible as $\operatorname{Spec} X = V(x) \cup V(y)$.
$X$ is irreducible does not imply $t(X)$ is irreducible: Just take $X$ to be the spectrum of any integral domain with exactly two maximal ideals.
The claim is true if $t(X)$ is dense in $X$, which is the case for example when $X$ is locally of finite type over $k$.