Is it true that if $X$ is connected, then for every nonempty proper subset $A$ of $X$, we have $\mathbf{Bd} \ne \emptyset$ Is it true that if $X$ is connected, then for every nonempty proper subset $A$ of $X$, we have $\mathbf{Bd} \ne \emptyset$? Does the converse hold?
I start by trying to understand what $\mathbf{Bd}$ means in "formal terms" (i.e. $\bar{A} \cap \overline{X-A}$ is the 'boundary' of $A \in X$)?
This means I'm supposed to figure out if the boundary of subset of a connected space is empty or not, right? I think that if a set is connected then it contains no partition into two sets, so there would exist points that are both in $A$ and in the complement of $A$ in $X$. This would be an informal definition of a boundary. (A boundary contains points in a set $A$ and points outside of it, right?) 
What I mean is, if $X$ is connected, and $A \subset X$ is connected, then $A$ could have a boundary which means that technically, $X$ cannot be partitioned into $A$ and $A$-complement without making the boundary of $A$ an empty set.
Right?
 A: Be careful: any set with more than one element can be partitioned into two non-empty sets; a connected set cannot be partitioned into two non-empty relatively open sets. And in your last paragraph it’s the closures of $A$ and $X\setminus A$ whose disjointness makes the boundary of $A$ empty.
The easiest way to prove the result is to prove the contrapositive. Suppose that $A$ is a non-empty proper subset of a space $X$, and suppose further that $\operatorname{bdry}A=\varnothing$. As you say, $\operatorname{bdry}A=(\operatorname{cl}A)\cap\operatorname{cl}(X\setminus A)$, so $\operatorname{cl}A$ and $\operatorname{cl}(X\setminus A)$ are disjoint closed sets whose union is $X$. Each of them is therefore the complement of a closed set, so each is open. Moreover, $\operatorname{cl}A\supseteq A\ne\varnothing$ and $\operatorname{cl}(X\setminus A)\supseteq X\setminus A\ne\varnothing$, so $\{\operatorname{cl}A,\operatorname{cl}(X\setminus A)\}$ is a partition of $X$ into non-empty open sets, and $X$ therefore is not connected.
A: Your intuition about their equivalence is spot on. The following conditions can be shown to be equivalent:


*

*$X$ is connected.

*If $A$ is a subset of $X$ that is both closed and open, then $A=\emptyset$ or $A=X$.


Observing that a subset of $X$ has empty boundary if and only if it is both closed and open, the equivalence follows immediately.
It's worth noting that any connected space with more than one point can be partitioned into two disjoint non-empty sets. However, it cannot be partitioned in this way with both sets open (or both sets closed, which is the same thing).
A: There are a few issues with your question, I'm going to address them.
First, a connected set cannot be partitioned into two open sets. Otherwise only singletons would be connected. Equivalently, it is connected iff its only clopen sets are the empty set and the whole space.
Second, the boundary of $A$ does not necessarily contain "points in $A$ and points not in $A$". For example, the boundary of $A = (0,1)$ is $\{0, 1\}$ and does not contain any points in $A$. If $A = [0,1]$ then every point of its boundary is in $A$.
Your last paragraph makes no sense to me, sorry.
There is an alternative definition of the boundary of $A$: it's the set $\bar A \setminus \mathring A$. This means that a point is in the boundary of $A$ if:


*

*It is in its closure, and

*It is not in its interior.


So now suppose $A \subset X$ is a proper nonempty subset of $X$. Assume that its boundary is empty. This means $\bar A \setminus \mathring A = \emptyset$. Therefore $\bar A = \mathring A$, and this set is also equal to $A$ because $\mathring A \subset A \subset \bar A$. But! By defintion, $\bar A$ is closed, and $\mathring A$ is open. Therefore the set $A = \bar A = \mathring A$ is clopen. This is a contradiction, because we assumed $A$ was a proper nonempty subset of $X$, and $X$ is connected.
The converse is also true. If every nonempty proper subset $A$ has nonempty boundary, then it can't be clopen, because otherwise $A = \bar A$ (closed) and $A = \mathring A$ (open), and therefore the boundary is empty. This is a contradiction.
