I am going to ask the question on general topology but the question is quite popular and I've solved it by myself but one moment seems to me very confusing. Please do not duplicate this question!
Let $A\subset X$. Let $C$ be a connected subspace of $X$ that intersects both $A$ and $X-A$. Prove that $C$ intersects the boundary of $A$, $\partial A$.
Proof: Suppose by contrary that $C$ does not intersect the $\partial A$, i.e. $C\cap \partial A=\varnothing$. It means that $C\subseteq X-\partial A$. But $X=\partial A\sqcup \text{Int}(A)\sqcup\text{Ext}(A)$ which implies that $C\subseteq \text{Int}(A)\sqcup\text{Ext}(A)$.
Hence $C=(C\cap \text{Int}(A))\sqcup(C\cap\text{Ext}(A))$. Since both of these sets are open in $C$ and since $C$ is connected then it one of them should be empty. WLOG we can assume that $C\cap \text{Int}(A)=\varnothing$. Hence $C\subseteq X-\text{Int}(A)=\overline{X-A}$. Since $C\cap A\neq \varnothing$ then there is some $p$ in this intersection. Then $p\in C\subset \overline{X-A}$ and $p\in A\subseteq \overline{A}$ so $p\in \overline{A}\cap \overline{X-A}$ and the last intersection by definition is the $\partial A$. Combining all this I got that $p\in \partial A\cap C$ which is contradiction.
The solution in my opinion looks completely correct but one moment seems to me a bit confusing: More precisely, I have never used the fact that $C\cap(X-A)\neq \varnothing$. Right?
Can anyone explain it to me please?
EDIT: However, after some time after creating this topic I have realized that we will be using the condition $C\cap (X-A)\neq \varnothing$ in the case $C\cap \text{Ext}(A)=\varnothing$, right? Am I correct?
Would be very thankful for useful feedback!
Your proof is correct. That you did not use $C\cap (X-A)\neq \varnothing$ is due to "WLOG we can assume that $C\cap \text{Int}(A)=\varnothing$". The other case $C\cap \text{Ext}(A)=\varnothing$ can of course be treated similarly, but must not be omitted. Considering this case you need $C\cap A\neq \varnothing$.
By the way, you can simplify your argument. If $C\cap \text{Int}(A)=\varnothing$, then necessarily $C \cap \partial A \ne \emptyset$ because $C \cap A \ne \emptyset$.