Can someone please verify my proof or offer suggestions for improvement?
Definition/Notation: The boundary of $A$, denoted by $\operatorname{Bd}(A)$, equals $\overline{A} \cap \overline{X-A}$.
Let $A \subseteq X$. Show that if $C$ is a connected subspace of $X$ that intersects both $A$ and $X-A$, then $C$ intersects $\operatorname{Bd}(A)$.
Suppose $C$ does not intersect $\operatorname{Bd}(A)$. Then, for each $x \in C$, it must be the case that either $x \notin \overline{A}$ or $x \notin \overline{X-A}$. If $x \notin \overline{A}$, pick an open set $U_x$ in $X$ such that $x \in U_x \subseteq X-A$, and set $V_x = \varnothing$. Otherwise, it must be the case that $x \notin \overline{X-A}$. In this case, pick an open set $V_x$ in $X$ such that $x \in V_x \subseteq A$ and set $U_x = \varnothing$.
Now, note that the sets $C \cap \bigcup_{x \in C} U_x$ and $C \cap \bigcup_{x \in C} V_x$ are open in $C$ under the subspace topology and are also nonempty (since there exists an $x \in C$ belonging to $\overline{A}$ and a $y \in C$ belonging to $\overline{X-A}$, otherwise one of $C \cap \overline{A}$ and $C \cap \overline{X-A}$ is nonempty, which implies that one of $C \cap A$ and $C \cap (X-A)$ is empty, a contradiction.)
By the choice of each $U_x$ and $V_x$, it can easily be seen that that $C \cap \bigcup_{x \in C} U_x$ and $C \cap \bigcup_{x \in C} V_x$ are disjoint, and their union equals $C$. This implies that $C$ is not connected, a contradiction. Therefore, it must be the case that $\operatorname{Bd}(A)$ intersects $C$.
Best Answer
If $C\cap\mathrm{Bd}(A)=\varnothing$ then $C=(C\cap\mathrm{int}(A))\cup(C\cap\mathrm{int}(X\setminus A))$ is a decomposition of $C$ as the union of two disjoint relatively open subsets. By hypothesis, both $C\cap\mathrm{int}(A)$ and $C\cap\mathrm{int}(X\setminus A)$ are nonempty, which contradicts the connectedness of $C$.