Simple argument to prove connectedness/disconnectednes of $\mathbb R^n\setminus \mathbb R^m$. I thought about the standard fact from topology which says that 
$\mathbb R^{n}\setminus \mathbb R^m$, where $\mathbb R^n$, $\mathbb R^m$ have standard Euclidean topologies and $$\mathbb R^m=\{(x_1,\ldots,x_m,0,\ldots,0)\in \mathbb R^n\mid x_i\in \mathbb R\}$$ is 


*

*path-connected in case $n-m\geq 2$

*divides into two components in case $n-m=1$ (it is clear that in case $n=m$ we get empty set which is out of interests here.)



What is most suitable pure topological (simple to follow, but rigorous) argument to show these  facts in first topology course? 

I mean the argument suitable if we, for instance, are not assuming neither knowledge of dimension theory nor linear spaces staff as orthogonal sub-spaces and etc.   
Good reference will be appreciated as well. 
 A: Following Mirko's suggestion in the comment, I think the easiest way to go here is to prove the statements directly. For $n-m \ge 2$ you can pick $2$ generic points and exhibit a path connecting them explicitly. For $n-m=1$ you can show that any path in $\mathbb{R}^n$ connecting the two components will have to pass through $\mathbb{R}^m$ by the intermediate value theorem.
A: Name $A =\mathbb R^{n}\setminus \mathbb R^m$
Case $n-m=1$
You have $A = A^+ \cup A^-$  and $A^+ \cap A^- = \emptyset$ where
$$\begin{cases}
A^+ = \{x \in A \mid x_n >0\} = \varphi^{-1}(\mathbb R_+)\\
A^- = \{x \in A \mid x_n <0\} = \varphi^{-1}(\mathbb R_-)
\end{cases}
$$
with $\varphi$ the linear form $\varphi: (x_1, \dots, x_n) \mapsto x_n$. Both $A^+,A^-$ are convex hence path connected (and connected).
Case $n-m \ge 2$
Consider two points $P_1=(x_1, \dots, x_n)$ and $P_2=(y_1, \dots, y_n)$ in $A$. $(x_{n-1}, x_n)$ and $(y_{n-1},y_n)$ are two points in $D = \mathbb R^2 \setminus \{(0,0)\}$. Those two points can be connected with a path composed of at most two line segments in $D$: $[P_1,U] \cup [U,P_2]$ with $U = (u_1,u_2) \in D$.
The path consisting of the two line segments
$$\begin{cases}
[(x_1, \dots , x_n),(0, \dots, 0, u_1,u_2)]\\
[(0, \dots, 0, u_1,u_2)],(y_1, \dots , y_n)]
\end{cases}$$
joins $P_1$ to $P_2$ in $A$.
A: Here are some more details. For $n=m+1$ prove that $(0,...,0,1)$ and $(0,...,0,-1)$ are in different components. Note that $\Bbb R^n_+:=\{(x_1,...,x_n):x_n>0)\}$ and $\Bbb R^n_-:=\{(x_1,...,x_n):x_n<0)\}$ each is a nonempty closed-and-open subset in the relative topology of $\Bbb R^n\setminus \Bbb R^m$ (and $\Bbb R^n_-\cup\Bbb R^n_+=\Bbb R^n\setminus \Bbb R^m$). Alternatively (using that $\Bbb R^n\setminus \Bbb R^m$ is open, so connectedness would be the same as path-connectedness) use that by the Intermediate value theorem 
(as pointed out in the answer by quarague) any path between $(0,...,0,1)$ and $(0,...,0,-1)$ would have to contain a point with $x_n=0$ which belongs to $\Bbb R^m$. 
When $n\ge m+2$, take two arbitrary points $A=(x_1,...,x_{n-2},p,q)$ and $B=(y_1,...,y_{n-2},r,s)$ each in $\Bbb R^n\setminus \Bbb R^m$. There are some easy cases, e.g. if $pr>0$ then take the straight line segment between $A$ and $B$, it avoids $\Bbb R^m$. But, in all cases, we have $(p,q)\neq(0,0)\neq(r,s)$. The problem boils down to finding a path in $\Bbb R^2$ between $(p,q)$ and $(r,s)$ that avoids the origin. This would be easiest done with pictures, and sample (interesting) cases, e.g. if $p<0<r$ and $q=s=0$ then take a line segment from $(p,0)$ to $(p,1)$, then from 
$(p,1)$ to $(q,1)$, then from $(q,1)$ to $(q,0)$. You could incorporate (somewhere in the above procedure) also taking a line segment in $\Bbb R^m$ from 
$(x_1,...,x_{n-2})$ to $(y_1,...,y_{n-2})$. 
