Show that the non-zero reals are connected in every dimension other than $N=1$ Hopefully I tagged this question appropriately, and apologize if I haven't.
How do I show that $\mathbb{R}^N \backslash\left\{0\right\}$ is disconnected $\iff$ $N=1$?
I have the right to left side of it, that is:
If $A:= \ (-\infty,0)$, $B:= \ (0,\infty)$ and $S:= \ \mathbb{R}^1\backslash\left\{0\right\}$
Then $A\cup B\supset S$ and $A\cap B=\emptyset\subset S^c$
And, $A\cap S \neq \emptyset$ and $B\cap S \neq \emptyset$.
$\therefore \mathbb{R}^1\backslash \left\{0\right\}$ is disconnected.
The converse seems pretty trivial when drawing it out, but I am not too sure regarding where I should start for a rigorous proof of it. Any help would be appreciated.
$\mathbf{EDIT:}$
My apologies, I forgot to mention that we have yet to prove path connected implies connected. This proof seems the easiest, but is their any other slick ways to show it? Using basic topology?
 A: If $N>1$, then, for any two points $p,q\in\mathbb{R}^N$, you can find a path going from $p$ to $q$ which doesn't pass through $0$. Therefore $\mathbb{R}^N$ is path-connected, and so it is connected.
If you can't use the fact that path-connected $\implies$ connected, you can still use the idea from the previous paragraph. Let $A,B\subset\mathbb{R}^N\setminus\{0\}$ be two non-empty open subsets of $\mathbb{R}^N\setminus\{0\}$ such that $A\cap B=\emptyset$ and that $A\cup B=\mathbb{R}^N\setminus\{0\}$. Take $p\in A$ and $q\in B$. Let $\gamma\colon[0,1]\longrightarrow\mathbb{R}^N\setminus\{0\}$ be a continuous function such that $\gamma(0)=p$ and that $\gamma(1)=q$. Then $\gamma\bigl([0,1]\bigr)$ is connected, and, if we define $A^\star=A\cap\gamma\bigl([0,1]\bigr)$ and $B^\star=B\cap\gamma\bigl([0,1]\bigr)$, then $A^\star,B^\star\neq\emptyset$, $A^\star$ and $B^\star$ are open subsets of $\gamma\bigl([0,1]\bigr)$, $A^\star\cap B^\star=\emptyset$, and $A^\star\cup B^\star=\gamma\bigl([0,1]\bigr)$. This is impossible, since $\gamma\bigl([0,1]\bigr)$ is connected.
Of course, what I did was to prove that path-connected $\implies$ connected.
A: A geometric approach. For $x=(x_1,...,x_n)$ let $\|x\|=\sqrt {\sum_{j=1}^nx_j^2}\;.$ We have of course $\|x\|+\|y\|\geq \|x+y\|.$ And $\|\cdot \|$ is a strictly convex norm : If $x$ and $y$ are linearly independent then $\|x+y\|<\|x\|+\|y\|.$ 
Let $n>1.$ For brevity let $S=\Bbb R^n$ \ $\{0\}.$ For $x\in S$ let $B(x,\|x\|)=\{y\in S: \|y-x\|<\|x\| \}.$  
(i). If $y,y'\in B(x,\|x\|)$ then $\{ry+(1-r)y':r\in [0,1]\}\subset B(x,\|x\|).$
Because $\|x-(ry+(1-r)y'\|=\|r(x-y)+(1-r)y'\|\leq r\|x-y\|+(1-r)\|x-y'\|<\|x\|.$
(ii). If $\|x\|=\|x'\|\ne 0$ and $x'\ne \pm x$ then $B(x,\|x\|)\cap B(x'/\|x'\|)\ne \phi.$ 
Because $x,x'$ are linearly independent so $\|x-x'\|<\|x\|+\|(-x')\|=2\|x\|.$ Hence $\|x-(x+x')/2\| =\|(x-x')/2\|<\|x\|,$ so $(x+x')/2\in B(x,\|x\|).$ Interchanging $x$ with $x'$ we also have $(x+x')/2\in B(x',\|x'\|)$.
(iii).Let  $U, V$ be open disjoint subsets of $S$ with $U\ne \phi$ and $U\cup V=S.$ Take $x_0\in U.$ For brevity let $r=\|x_0\|.$  Then $B(x_0,\|x_0\|)\subset U.$ 
Proof: Let $y\in B(x,\|x|.$ Then $f:[0,1]\to T_y=\{tx+(1-t)y:t\in [0,1\},$ where $f(t)=t x_0+(1-t)y$, is continuous, and $[0,1]$ is connected, so $T_y$ is a connected space .  Since $T_y\cap U$ and $T_y\cap V$ are open disjoint subsets of the space $T_y$ whose union is $T_y,$ and since  $T_y\cap U\ne \phi$  (because $x_0\in T_y\cap U$) we have $T_y=T_y \cap U.$ So $y\in U.$  
(iv). If $\|x'\|=r=\|x_0\|$ and $x\ne \pm x_0$,  then by the method of (iii)  we have  $B(x',r)\subset U$ or $B(x',r)\subset V.$ But by (ii) $B(x_0,r)\cap B(x',r)\ne \phi =U\cap V ,$ so $B(x',r)\subset U.$
Now for dimension $n>1$ there exists $x'$ with $\|x'\|=\|x_0\|=r$ and $x'\ne \pm x_0.$ We have $B(x_0,r)\cap B(x',r)\ne \phi \ne B(-x_0,r)$ so $B(-x_0,r)$ cannot be a subset of $V$ . So $B(-x_0,r)\subset U.$ (This is where $n>1$ is used.)
Therefore the set $K= \{y\in S:\|y\|<2r\}=\cup \{B(x,r):\|x\|=r\}$ is a subset of $ U.$ 
Now if $z\in S$ with $\|z\|\geq 2r$  and if $z\in V$ then $B(z,\|z\|)\subset V.$ But $zr/\|z\|\in B(z,\|z\|)\cap K$ , a contradiction.
Therefore $U=S=\Bbb R^n$ \ $\{0\}.$ 
