Assume that $S \subset \mathbb{R}^n$ where $n>1$ then is the following connected? Assume that $S \subset  \mathbb{R}^n$ is a connected set where $n>1$ then let $A$ be a countable set of $S$. Is $S-A$ connected?
I was trying to solve the following :

show that there is no open subset of $\mathbb{R}$ homemorphic to an open subset of $\mathbb{R}^2$.

My attempt has been in the following way. Assume that there is a set $U \subset \mathbb{R}$ such that $U$ is open in $\mathbb{R}$. Now we see that $U$ can be written as the countable union of disjoint open intervals say $J_1,\cdots,J_n$ .
Let $f$ be a homeomorphism between $U$  and $V$ an open subset of $\mathbb{R}^2$. $f(J_1)=V'$ is connected as $J_1$ is an open interval.
Is $V' - \mathbb{Q^2}$ connected in $\mathbb{R}^2$ ? Then by the property of homeomorphism $f^{-1}(V' - \mathbb{Q^2}) = J_1 - A'$ where $A'$ is a countable set must be connected. I am trying to arrive at a contradiction.
Any hint on my first and second question would be appreciated.
 A: As pointed out in the comment, you probably want $S$ to be open, which is what you need in your proof. Also you only need to remove one point to disconnect an open interval. However I realized that I don't know a much quicker way to prove that $S$ with one point deleted is connected than proving it for $S$ with a countable set deleted, so let me prove the latter.
Suppose $Z\subseteq S$ is countable and $x,y\in S\setminus Z$. Consider the set $\{w\in S\mid w\text{ can be connected with $x$ by a finite number of straight segments in $S$}\}$. One can show that it is both open and closed in $S$, so it must be equal to $S$; in particular $y$ belongs to the set. Suppose there are points $u_1,...,u_n$ s.t. the segments $xu_1,u_1u_2,...,u_{n-1}u_n,u_ny$ are all contained in $S$. Let $l$ be any straight segment passing through $u_1$ and not on the line $xu_1$. If $u\in l$ is close enough to $u_1$, then the segments $xu,uu_2$ are in $S$. $xu\cap Z\neq\emptyset$ for only countably many $u$, so there exists some $u$ such that $xu\cap Z=\emptyset$, and we change $u_1$ to $u$. Then we change $u_2$ to another point, etc.
Alternatively, we can use the fact that $S$ is a connected $n$-dimensional manifold, so any two points $x,y$ are contained in an open set homeomorphic to $\mathbb{R}^n$ (this needs a proof, which is similar to the above argument). That reduces the problem to the case $S=\mathbb{R}^n$, which is slightly easier: just consider an uncountably family of disjoint curves from $x$ to $y$, and some of them must avoid $Z$.
