A connected set that must intersect the boundary of an open set given certain conditions Suppose that $C$ is a connected subset of $X$, and $U$ an open set in $X$. Also, we have that $C$ contains a point in $X$ and a point not in $X$. Then how do we show that the following is true: 
$$C \cap \partial U \neq \varnothing?$$  
My second question is that what is a good example in $\mathbb{R}^2$ of a connected set $C$ with an open set (disk) $U$ such that $C$ contains a point inside $U$ and a point in the complement of cl$(U)$, i.e. the closure of $U$, and some component of $U \cap C$ misses $\partial U$?  
For the first question, right now the only thing that comes to mind is a proposition that states for $A \subset X$, if $C$ is a connected subspace of $X$ such that 
$$C \cap A \neq \varnothing, C \cap (X-A) \neq \varnothing,$$ 
then $C \cap \partial A \neq \varnothing$. In this case, however, we don't know if $A$ is open, and nowhere is it mentioned that $C$ contains a point in $X$ and a point not in $X$. So how should we proceed? I would also really appreciate some guidance on the second question for a suitable example and why it works. 
 A: I assume that you meant to say that $C$ contains a point of $U$ and a point not in $U$.
HINT: Let $V=X\setminus\operatorname{cl}U$. Show that if $C\cap\operatorname{bdry}U=\varnothing$, then $U\cap C$ and $V\cap C$ form a separation of $C$, contradicting the assumption that $C$ is connected.
For the second question practically any example will work. For instance, let
$$U=\big\{\langle x,y\rangle\in\Bbb R^2:x^2+y^2<1\big\}\;,$$
the open unit disk centred at the origin, and let $C$ be the $x$-axis. The boundary of $U$ is the unit circle, and the only component of $U\cap C$ is $\big\{\langle x,0\rangle:-1<x<1\big\}$.
A: I will assume that you meant "another point not in $U$", I will call this point $y$. If $y \in  \partial U$ then we are done. Therefore, assume $y \notin ∂U$ and assume $C∩∂U≠ \varnothing$ . Now Let $V$ be the interior of the complement of $U$. Now we have $x \in C ∩ U$ and $y \in C ∩ V$ since $C$ is a subset of the union of the open disjoint sets $U$ and $V$ (because $C∩∂U≠ \varnothing$), thus we get a contradiction because $C$ is connected.
A: I would suggest to do as follows: Take A=int(U)∩C and B=int(complement of U)∩C.  We know that A∩B = ∅.  Since C is not a subset of A, c is not a subset of A∪B. Thus C≠A∪B. By taking the intersection of both sides with ∂U we can show the latter.
