# Any two points contained in a connected subset implies the boundary of a proper set is non-empty

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?

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The assumption exactly means that $X$ is connected. One direction is clear, as we can always take $X$ for any pair of points, and for the other direction, fix on point $p$ and let $C_x$ be a connected subset of $X$ that contains $p$ and $x$ for each $x \neq p$. Then $\Cup_{x \neq p} C_x$ is connected (standard theorem) and equals $X$ obviously.
The fact that all non-empty proper subsets of $X$ have non-empty boundary also is equivalent to $X$ being connected: if $A, X \setminus A$ is a separation of $X$ then the boundary of $A$ is empty (the boundary equals the closure minus the interior, so a set is clopen off its boundary is empty). And if $A$ is proper, non-empty and has empty boundary, $A$ and $X \setminus A$ form a separation of $X$.
So both are just (well-known) characterizations of the connectedness of $X$.