We will use without further mention that $X_0$ is quasi-compact (Topology, Lemma \ref{topology-lemma-closed-points-quasi-compact}) and $\pi_0(X)$ is profinite (Topology, Lemma \ref{topology-lemma-spectral-pi0}). Picture $$ \xymatrix{ X_0 \ar[rd]_f \ar[r] & X \ar[d]^\pi \\ & \pi_0(X) } $$ If (2) holds, the continuous bijective map $f : X_0 \to \pi_0(X)$ is a homeomorphism by Topology, Lemma \ref{topology-lemma-bijective-map}. Given an open covering $X = \bigcup U_i$, we get an open covering $\pi_0(X) = \bigcup f(X_0 \cap U_i)$. By Topology, Lemma \ref{topology-lemma-profinite-refine-open-covering} we can find a finite open covering of the form $\pi_0(X) = \coprod V_j$ which refines this covering. Since $X_0 \to \pi_0(X)$ is bijective each connected component of $X$ has a unique closed point, whence is equal to the set of points specializing to this closed point. Hence $\pi^{-1}(V_j)$ is the set of points specializing to the points of $f^{-1}(V_j)$. Now, if $f^{-1}(V_j) \subset X_0 \cap U_i \subset U_i$, then it follows that $\pi^{-1}(V_j) \subset U_i$ (because the open set $U_i$ is closed under generalizations). In this way we see that the open covering $X = \coprod \pi^{-1}(V_j)$ refines the covering we started out with. In this way we see that (2) implies (1). \medskip\noindent Assume (1). Let $x, y \in X$ be closed points. Then we have the open covering $X = (X \setminus \{x\}) \cup (X \setminus \{y\})$. It follows from (1) that there exists a disjoint union decomposition $X = U \amalg V$ with $U$ and $V$ open (and closed) and $x \in U$ and $y \in V$. In particular we see that every connected component of $X$ has at most one closed point. By Topology, Lemma \ref{topology-lemma-quasi-compact-closed-point} every connected component (being closed) also does have a closed point. Thus $X_0 \to \pi_0(X)$ is bijective. In this way we see that (1) implies (2). \medskip\noindent Assume $X_0$ is closed in $X$ and every point specializes to a unique point of $X_0$. Then $X_0$ is a spectral space (Topology, Lemma \ref{topology-lemma-spectral-sub}) consisting of closed points, hence profinite (Topology, Lemma \ref{topology-lemma-characterize-profinite-spectral}). Let $x, y \in X_0$ be distinct. By Topology, Lemma \ref{topology-lemma-profinite-refine-open-covering} we can find a disjoint union decomposition $X_0 = U_0 \amalg V_0$ with $U_0$ and $V_0$ open and closed and $x \in U_0$ and $y \in V_0$. Let $U \subset X$, resp.\ $V \subset X$ be the set of points specializing to $U_0$, resp.\ $V_0$. Observe that $X = U \amalg V$. By Topology, Lemma \ref{topology-lemma-make-spectral-space} we see that $U$ is an intersection of quasi-compact open subsets. Hence $U$ is closed in the constructible topology. Since $U$ is closed under specialization, we see that $U$ is closed by Topology, Lemma \ref{topology-lemma-constructible-stable-specialization-closed}. By symmetry $V$ is closed and hence $U$ and $V$ are both open and closed. This proves that $x, y$ are not in the same connected component of $X$. In other words, $X_0 \to \pi_0(X)$ is injective. The map is also surjective by Topology, Lemma \ref{topology-lemma-quasi-compact-closed-point} and the fact that connected components are closed. In this way we see that the final condition implies (2).