Proof verification: $\mathbb{R^n}$ and $\mathbb{R}$ are not homeomorphic Question:

Show $\mathbb{R^n}$ and $\mathbb{R}$ are not homeomorphic if $n>1$.

My Solution:
Let $p:\mathbb{R}\rightarrow \mathbb{R^n}$ is a quotient map.
$p(\{0\})=\{0^n\}$ is a closed map.($0^n=(0,0,\dots,0)=0\times 0\times \dots\times 0$)
Now, $\mathbb{R}-\{0\}$ is separated, but $\mathbb{R^n}-0^n$ is connected. Let the separations are $A$ and $B$.
Let $q:A\cup B\rightarrow p(A\cup B)$ is a quotient map.
Let, $q(A\cup B)=\mathbb{R^n}-\{0^n\}$. Then $q(A\cap B)=q(\emptyset)=q(A)\cap q(B)=C$. Now $C$ is a open set. But $q(\emptyset)$ is both open and closed, then $q(\emptyset)=\emptyset$. Then $q(A)$ and $p(B)$ are separations, which contradicts that $p$ is a quotient map.
Let $q(A\cup B)=\mathbb{R^n}$. Then $p(0)\in q(A\cup B)$. Then $0\in A\cup B$. which contradicts our assumption.
Is my argument ok?
Edit:
Thanks to Randall for pointing out my solution is too broad. So I will be grateful if someone give a shorter solution using Connctedness (avoiding path-connectedness would be better!)
 A: Assume to the contrary that $h: \mathbb{R} \to \mathbb{R}^n$ is a homeomorphism and choose a point $p \in \mathbb{R}$.  Then $h$ induces another homeomorphism $h' : \mathbb{R} -\{p\} \to \mathbb{R}^n-\{h(p)\}$.  But the codomain is (path) connected while the domain is not, a contradiction. 
If you want to do this using only connectivity, you need to prove that $\mathbb{R}^n- \{\mathbf{0}\}$ is connected when $n > 1$.  (The easiest way to do this is to prove it is path connected, sorry.)
You can be cheap and hide the path connected argument inside another.  Let $X = \mathbb{R}^n-\{\mathbf{0}\}$ out of laziness, and assume that $X$ is disconnected with separation $X = A \cup B$ (open, disjoint, non-empty).  Given points $a \in A$ and $b \in B$ argue that there is a path $\alpha: [0,1] \to X$ between them.  But the image of $\alpha$ must be connected, yet it contains $a$ and $b$, a contradiction.  (I feel dirty typing this out.)  
Edit: wait, here's a potentially better idea for your needs. Do you believe that $\mathbb{R}^n-\{\mathbf{0}\}$ is homeo to $S^{n-1} \times \mathbb{R}$?  If so, then the assumption $n > 1$ shows that the latter is connected (a product of connected spaces), which is what you wanted.
