Assume that there exists a separation $A \ \cup \ B \ = \ X$. It would follow that each $p^-1(\{y\})$ would have to be contained completely in either $A$, or $B$ (not both, or else $A \ \cap \ p^{-1}(\{y\})$ and $B \ \cap \ p^-1(\{y\})$ would separate $p^{-1}(\{y\})$). Thus, $A$ and $B$ are saturated open sets. Since a quotient map maps saturated open sets to open sets, it follows that $p(A)$ and $p(B)$ are open set whose union is $p(X) \ = \ Y$ (from the surjectivity of $p$). Assume that $p(A) \ \cap \ p(B) \ \neq \ \emptyset$. It would follow that there exists some $y$ such that $p(a) \ = \ p(b) \ = \ y$, for $a \ \in \ A$ and $b \ \in \ B$. It would follow that $p^{-1}(\{y\})$ would intersect both $A$ and $B$, which can't be the case. Thus, $p(A)$ and $p(B)$ are disjoint, and therefore form a separation of $Y$. This is a contradiction, so $X$ must be connected.

Assume that $C$ does not intersect $\text{Bd} \ A$. This implies that there exists no point of $C$ that is in both $\bar{A}$ and $\overline{X \ - \ A}$. Consider some point $x \ \in \ C$. If $x \ \in \ C \ \cap \ A$, it follows that there exists at least one neighbourhood of $x$ that does not intersect $X \ - \ A$ (or else $x \ \in \ \overline{X \ - \ A}$). Similarly, if $x \ \in \ C \ \cap (X \ - \ A)$, it follows that there exists at least one neighbourhood of $x$ that does not intersect $A$ (or else $x \ \in \ \bar{A}$). We call such open sets $U_x$ for each $x$. For each $x \ \in \ C$, consider the intersection $U_x \ \cap \ C$. Each of these sets are open in $C$. Now, consider: $$U_{A} \ = \ \displaystyle\bigcup_{x} \ U_x \ \cap \ C \ \ \ \ \ \ x \ \in \ C \ \cap \ A$$ $$U_{X \ - \ A} \ = \ \displaystyle\bigcup_{x} \ U_x \ \cap \ C \ \ \ \ \ \ x \ \in \ C \ \cap \ X \ - \ A$$ Clearly, $U_A$ and $U_{X \ - \ A}$ form a separation of $C$, contradicting the fact that it is connected. Thus, $C$ must intersect $\text{Bd} \ A$.