Let $R$ be a valuation ring of a field $K$. Let $T=\operatorname{Spec} R$ and let $U=\operatorname{Spec} K$.In Harthorne Lemma II.4.4 he proves that given a scheme $X$, a morphism of $T$ to $X$ is equivalent to giving two points $x_0,x_1$ in $X$, with $x_0 \in \overline{\{x_1\}}$ and an inclusion $k(x_1)\subseteq K$, such that $R$ dominates the local ring $\mathcal{O}$ of $x_0$ on the subscheme $Z= \bar{\{x_1\}}$ of $X$ with its reduced induced structure.
I am trying to understand the following step: write $t_0 = \mathfrak{m}_R$ and $t_1 = (0)$. Given a morphism $T \to X$ we let $x_0$ and $x_1$ be the image of $t_0$ and $t_1$ respectively. He then says "Since $T$ is reduced, the morphism $T \to X$ factors through $Z$ (Ex 3.11)"
The Ex 3.11 he is referring to, which I think he means 3.11 (c), requires i) the topological space of $T$ and $Z$ to be homeomorphic and ii) $T \to X$ has to be a closed immersion.
Any explanation on these two points would be appreciated. Thank you.