Are there nice circumstances under which connectedness of interior and boundary imply connectedness? Munkres problem 24.11: If $A$ is a connected subspace of $X$, does it follow that $\operatorname{Int}A$ and $\operatorname{Bd}A$ are connected? Does the converse hold?
I've answered these questions, but I'm wondering if the converse might hold if $X$ satisfies some reasonable conditions, like being normal and connected, or perhaps locally connected, and assuming that neither the interior nor the boundary is empty.
 A: The converse fails to hold even in very pleasant topological spaces, like the reals. Of course, $\mathbb R$ is perfectly normal, connected, etc. However, even though $\mathbb Q \subseteq \mathbb R$ has both its boundary (namely, $\mathbb R$ itself) and its interior (the empty interior) connected, the set $\mathbb Q$ itself is disconnected.

Tweak to get nonempty interior: The OP desired a set $A$ with nonempty interior in the comments. For this, take the disconnected set $A = (-\infty, 0] \cup \mathbb Q_{\geqslant 0}$.  However, its interior $\operatorname{Int} A = (-\infty, 0)$ and its boundary $\operatorname{Bd} A = [0, \infty)$ are both nonempty and connected.
A: $\DeclareMathOperator{\Int}{Int}\DeclareMathOperator{\Bd}{Bd}$
The most interesting thing I came up with, which may be a special case of something more interesting, is that if $X$ is connected, $A\subset X$, and $\Int A$ and $\Bd A$ are
connected, then $\bar{A}$ is connected:
Suppose that $B$ and $C$ form a separation of $\bar{A}$. Since
$\bar{A}$ is closed, $B$ and $C$ are closed in $X$. Since $\bar{A}=\Int A\cup\Bd A$
and the interior and boundary are connected, we can assume without
loss of generality that $\Int A\subset B$ and $\Bd A\subset C$,
so in fact $\Int A=B$. Furthermore, since $\Int A$ is connected,
so is its closure, and $\overline{\Int A}\subset\bar{A}$, so $\overline{\Int A}=\Int A$.
But this contradicts the fact that $X$ is connected.
