Claim:
$ int(A\cup B) = int(A) \cup int(B) \iff \partial A \cap \partial B \subset \partial (A \cup B) $
Proof:
Since you are more intrested in the ($\Leftarrow$) part of the claim I will first prove that. For both parts of the proof, remeber that interior, exterior and boundary of a set is a partitioning of the whole space i.e. $int(A) \cup \partial A \cup ext(A) = X$ where all three sets are disjoint. Also, interior and exterior are open sets.
($\Leftarrow$)
As you have said $int(A) \cup int(B) \subset int(A\cup B)$. So, I will only focus on proving $int(A\cup B) \subset int(A) \cup int(B)$.
Let $x \in int (A \cup B)$. If $x \in int(A)$ or $x \in int(B)$ then we are done. So, assume that $x \notin int(A)$ and $x \notin int(B)$.
By hypothesis, if $x \in \partial A \cap \partial B$ then $x \in \partial(A\cup B)$ but interior and boundary of $(A \cup B)$ are disjoint which is a contradiction. So, $x \notin \partial A \cap \partial B$.
Without loss of generality, assume that $x \notin \partial B$. Then, $x \in ext(B)$ and either $x \in \partial A$ or $x \in ext(A)$.
Case 1: $x \in \partial A$,
Since $x \in ext(B)$, there is a neighborhood $V \subset B^C$ of x. Since $x \in int(A \cup B)$ there is a neighborhood $U \subset (A\cup B)$ of x. Then,
$U \cap V$ is a neighborhood of $x$ in $X$. Note that $(U \cap V) \subset (A\cup B) \cap B^C \subset A$. But since $x \in \partial A$, any neighborhood of $x$ must intersect non-trivially with $A^C$. Hence, it is a contradiction. So, $x \notin \partial A$.
Case 2: $x \in ext(A)$,
$x \in int(A \cup B) \subset A \cup B \subset \bar A \cup \bar B = (int(A) \cup
\partial A) \cup (int(B) \cup \partial B)$. However $x \in ext(A)$ and $x \in ext(B)$ implying that $x \notin int(A) \cup \partial A$ and
$x \notin int(B) \cup \partial B$ which is a contradiction.
Therefore, $x \in int(A)$ or $x \in int(B)$ i.e. $x \in int(A) \cup int (B)$.
$ (\Rightarrow) $ Let $x \in \partial A \cap \partial B$.
So, $x \notin int(A) $ and $ x \notin int(B)$ implying $x \notin int(A) \cup int(B) = int(A\cup B)$.
Thus, $x \in \partial (A \cup B) \cup ext(A\cup B)$.
Assume $x \in ext(A \cup B)$ then there is a neighborhood $P \subset (A \cup B)^C $ of $x$ (where $A^C$ means the complement for set $A$).
But $(A \cup B)^C \subset A^C$. That is, we have a neighborhood of $x$ which is contained in $A^C$ meaning $ x \notin \partial A$ which is a contradiction.
Hence, $x \in \partial(A\cup B)$.
This is too long to be a comment and is currently only a partial answer.
Following the idea that the existence of an open, totally disconnected set should relate to the connectedness of the space, I read about ways a space can be connected. A locally connected space is one where every point in a open set has a connected, open neighborhood and felt most relevant. I wasn't able to show that being locally connected was a sufficient condition for the lack of an open, totally disconnected set. So, I ended up adding connected back as a requirement (and T1 ended up being helpful during the proof) which led to,
Let $(X,\tau)$ be a locally connected, connected, T1 space that is not the topology on a one point set. Then, $(X, \tau)$ is not weakly disconnected.
Proof:
The proof strategy will be to use proof by contrapositive. Let $(X, \tau)$ be a weakly disconnected space. Let $U$ be an open, non-empty totally disconnected set. Let $(U, \tau_{U})$ be the subspace topology of $U$. As $U$ is totally disconnected and non-empty, its connected components are singletons.
Here, we'll split the proof into two cases. Either all of its connected components are open or they there exists a connected component that is not open.
If there exists a connected component that is not open then its connected components are not all open which means it is not locally connected. As locally connected is hereditary for open subspaces, $(X,\tau)$, can not be locally connected.
If there exists a connected component that is open, call that component $A$. As $U$ is totally disconnected, $A$ must be a singleton. Since $U$ is open, any open set in $U$ is also open in $X$. This means $A$ is an open singleton in $X$. As $X$ is a T1 space, singletons are closed making $A$ a clopen set. Since $X$ is not a one point set, $A$ is non-trivial clopen set. This means $(X,\tau)$ is not connected.
In either case, $(X, \tau)$ can not have the three properties of being locally connected, connected, and T1. This completes the proof.
The issue is I'm not sure if those three properties are a necessary condition, so I still lack a nice way of describing all spaces that lack an open, totally disconnected set. While my current proof strategy uses the fact the space is T1, I'm skeptical that there isn't an alternate way to drop that condition.
Best Answer
For a point $x$ to be in the interior of $A\subset X$ (with respect to $X$) means that all the point of $X$ close enough to $x$ are also in $A$ (ie $A$ contains a neighborhood of $x$).
Now, for $x$ to be in $\operatorname{Int}_Y(A)$, you need all the points of $Y$, close enough of $x$, to be in $A$. Since $Y\subset X$, there are less points that we need to consider, so this is a weaker condition.