connected neighborhood of $0$ must be in unit ball 

In topological vector space $C^{n}$, let $B$ be the open unit ball and $S$ be the unit sphere. Suppose $E$ is a connected neighborhood of $0$ disjoint from $S$. Prove that $E \subset B$


This fact seems intuitively natural, but I can't prove it formally using lemmas of connectedness. Can any one help me? Thanks
 A: It is somehow Jordan's Theorem about closed curves of $S^2$ that separate the sphere into two disjoint connected components. I'll make a try to this one applying somehow Jordan's Theorem.
(All curves here are continuous)
We have that $E\cap B\neq \emptyset$ thus it would be logical to assume(we want to prove by contradiction) there is a $x\in E:x\notin B$. Because $E\cap B\neq \emptyset$ we have that $x\notin S$ too, so it is outside of $S$.We also want $γ$ be the path that unites $x$ with $0$(this exists because $E$ is connected). 
Now take a curve on $S$ that cuts it in half(something like the hemisphere). Then we get two disjoint connected components let's say $C^1_1,C^2_1$.(This stands due to Jordan's Theorem). Then we have that $\gamma$ unites $x$ with $0$ only by passing through $C^1_1$(just choose one of the two) because if it passed through $C^2_2$ then we wouldn't have two different connected components because $E$ is connected.Do the same now for a stricky different hemisphere that cuts $S$ in $C^1_2,C^2_2$. Same here, $\gamma$ passes through one and only one,let's say $C^1_2$.
Now what we have achieved so far is to build a closed space in $S$ which is the space that is  the intersection of $C^1_1$ and $C^1_2$. We continue this prossess infinitly. So we build closed sets(which are bounded and so compact) $F_{i},i\in \Bbb N: F_1\supset F_2\supset F_3\supset...$ and hence $\cap_{i\in \Bbb N} F_i=${$y$} for $y\in S$. So we have $E$ looking something like a (let's say) bubble inside $B$ with a straight line ($\gamma$) that goes from $0$ to $x$ and passing through $y\in S$.
Now we use the neightbourhood thing. $E\in n_0=>0\in E^{o}$. Of course $x\notin E^{o}$. So $x$ unites with $0$ through the boundary of $E$. So $\gamma \in Bd(E)$ and $E$ is closed on the path $\gamma$ and locally to the point where $\gamma$ reaches the main body of $E^{o}$. This means that $Bd(E)\subset E$ in the points that i said before and because $Bd(E)\cap S\neq \emptyset=>E\cap S \neq \emptyset$.
