Let $X$ be a topological space and suppose any two points in $X$ are contained in a connected subset of $X$. Show that if $A \subset X$ is a proper subset then $\operatorname{bd}(A) \neq \emptyset$.
Well basically I cheated here, because I knew the exercise that if $C \subset X$ is connected, $A \subset X$ then if $C$ intersects $A$ and $X \setminus A$ then $C$ intersects $\operatorname{bd}(A)$.
Here's how I applied it:
Suppose, for sake of contradiction, that if $A \subset X$ is proper but that $\operatorname{bd}(A)=\emptyset$. Now pick $a \in A$ and $b \in X \setminus A$. By assumption there is a connected subset $C$ of $X$ such that $a \in C$ and $b \in C$. Then $C$ meets $A$ and $X \setminus A$ so the exercise applies and then $C$ intersects $\operatorname{bd}(A)$. This implies that $\operatorname{bd}(A)$ is non-empty, a contradiction.
Question: how to show this without using the exercise?