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.
For every index $\alpha \in A$ there is a canonical inclusion map $\sigma_\alpha :X_\alpha \to \coprod_{\alpha \in A} X_\alpha$ given by $x \mapsto (x,\alpha)$. A subset $V$ of $\coprod_{\alpha \in A} X_\alpha$ is declared as open if $\sigma_\alpha^{-1}(V)$ is open in $X_\alpha$ for every $\alpha \in A$. More precisely, we define a topology $\mathfrak{T}$ on $\coprod_{\alpha \in A} X_\alpha$ as follows:
$$
\mathfrak{T} = \left\{ V \subseteq \coprod_{\alpha \in A} X_\alpha : \sigma_\alpha^{-1}(V) \text{ is open in $X_\alpha$, for all $\alpha$} \right\}.
$$
As you pointed out, we then identify $X_\alpha$ with the set
$$
Y_\alpha := \{ (x,\alpha) : x \in X_\alpha\}.
$$
Then, $\sigma_\alpha^{-1}(V)$ can be viewed as the "intersection of $V$ with $X_\alpha$". Indeed, after identifying $X_\alpha$ with $Y_\alpha$ above, the pre-image $\sigma_\alpha^{-1}(V)$ asks which points in $V$ live in $X_\alpha$.
Best Answer
If your definition is that a space $X$ is disconnected iff it can be written as $U\cup V$, where $U$ and $V$ are disjoint, non-empty open sets, then your approach is exactly right. If $X$ is disconnected, let $U$ and $V$ be as above, and show that $X$ is homeomorphic to the disjoint union of $U$ and $V$. Depending on your definition of disjoint union of spaces, this may be completely trivial, and you can take the homeomorphism to be the identity map. At worst your definition of the disjoint union of $U$ and $V$ may be something like $(U\times\{0\})\cup(V\times\{1\})$, and there’s still a very natural homeomorphism between that and $X$, very closely related to the identity map.
For the other direction you have $X$ homeomorphic to $\bigsqcup_{i\in I}X_i$, a disjoint union of spaces $X_i$ for $i$ in some index set $I$ with at least two members. Let $h:\bigsqcup_{i\in I}X_i\to X$ be the homeomorphism. Pick $i_0\in I$, and let $U=h[X_{i_0}]$; what should $V$ be to give you a disconnection of $X$?