Your property is equivalent to complete normality.
Recall that subsets $A$ and $B$ of $X$ are said to be separated if $A\cap\operatorname{cl}B=\varnothing=B\cap\operatorname{cl}A$, and $X$ is completely normal if every pair of separated sets in $X$ have disjoint open nbhds. $X$ is $T_5$ if $X$ is completely normal and $T_1$.
Suppose that $X$ is completely normal, and let $Y$ be a subspace of $X$ that is not connected. Then there are open sets $U_0$ and $U_1$ in $X$ such that if $G_0=U_0\cap Y$ and $G_1=U_1\cap Y$, then $\{G_0,G_1\}$ is a partition of $Y$ into (non-empty) relatively open sets. $U_0$ is an open nbhd of $G_0$ disjoint from $G_1$, so $G_0\cap\operatorname{cl}_XG_1=\varnothing$. Similarly, $G_1\cap\operatorname{cl}_XG_0=\varnothing$, so $G_0$ and $H_0$ are separated subsets of $X$. $X$ is completely normal, so there are disjoint open sets $V_0$ and $V_1$ such that $G_0\subseteq V_0$ and $G_1\subseteq V_1$. This shows that every completely normal space has your property.
Conversely, suppose that $X$ has your property, and let $A$ and $B$ be separated sets in $X$. Then $A$ and $B$ are disjoint, non-empty, relatively closed subsets of $A\cup B$, which is therefore not connected. Since $X$ has your property, $A$ and $B$ have disjoint open nbhds in $X$, and $X$ is therefore completely normal.
By the way, complete normality is equivalent to hereditary normality: all subspaces are normal.
Note that every space with the indiscrete topology is vacuously completely normal and therefore has your property, so your property does not imply that the space is Hausdorff. If $X$ is $T_1$ and has your property, however, then singletons are closed, and $X$ must be Hausdorff.
Best Answer
No, it isn't: It is well-known, and easy to see, that a space is extremally disconnected, iff the closure of each open set is open again.
Let $U := \omega \times \{0 \}$. Then $U$ is open, $\overline{U} = U \cup \{x\}$, which is not open.