The hard part of this is to show that given $2n$ distinct points of $\Bbb R^{2}$ $x_1,\ldots,x_n,y_1,\ldots,y_n$ there exists a homeomorphism from $\Bbb R^{2}$ onto itself which sends $x_i$ to $y_i$ for all $i=1,\ldots,n$. This can be done using tubular neighborhoods as follows:
First, consider the set of all ellipses whose endpoints of their mayor axes are $x_1$ and $y_1$. Notice that if we take two distinct such ellipses $C_1$ and $C_2$, then $C_1\setminus\{x_1,y_1\}$ and $C_2\setminus\{x_1,y_1\}$ are disjoint, and thus as there infinitely many such ellipses there must exist one which does not contain any of the points $x_2,\ldots,x_n,y_2,\ldots,y_n$.
Let $C$ be such an ellipse. The tubular neighborhood theorem says that there exists some $\epsilon>0$ and a homeomorphism $\varphi:[-1,1]\times C\rightarrow \overline{D_\epsilon(C)}$ such that $\varphi(\{0\}\times C)=C$, and this also holds for any other positive $\delta<\epsilon$, where $D_\epsilon(C):=\{x \in\Bbb R^2:d(C,x)<\epsilon\}$.You can try to prove this by taking any $\epsilon>0$ less than half the length of the minor axis of the ellipse. Pick $\epsilon$ small enough so that $\overline{D_\epsilon(C)}$ does not contain any of the points $x_2,\ldots,x_n,y_2,\ldots,y_n$.
As $C$ is homeomorphic to a circle, it is easy to see that there is a homeomorphism $[-1,1]\times C\rightarrow [-1,1]\times C$ sending $(0,x_1)$ to $(0,y_1)$ and which is the identity on the boundary of $[-1,1]\times C$(Can you see why?). Thus there is a homeomorphism of $\overline{D_\epsilon(C)}$ onto itself which sends $x_1$ to $y_1$ and is the identity on $\partial \overline{D_\epsilon(C)}$. Hence this homeomorphism extends to a homeomorphism of $\Bbb R^2$.
Continuing this process by induction we get the desired homeomorphism of $\Bbb R^2$; at intermediate steps we pick the tubular nhoods so that they don't touch the other previous tubular nhoods.
By what we did above we know $\Bbb R^2\setminus\{n$ points $\}$ is homeomorphic to $\Bbb R^2\setminus\{(1,0),\ldots,(n,0)\}$.
There is a strong deformation retract of $\overline{D_1(0)}$ onto its boundary. Thus $\Bbb R^2\setminus\{(1,0),\ldots,(n,0)\}$ is homotopic to $\Bbb R^2\setminus(D_{1/2}((1,0))\cup\cdots\cup D_{1/2}((n,0)))$.
Now $\Bbb R^2\setminus(D_{1/2}((1,0))\cup\cdots\cup D_{1/2}((n,0)))$ is clearly homotopic to $X=S_{1/2}((1,0))\cup\cdots\cup S_{1/2}((n,0))\cup\{(x,0):x\in (-\infty,1/2]\cup[n+1/2,\infty)\}$, where $S_i(x)=\{y\in\Bbb R^2:||x-y||=i\}$. These last topological space is in turn clearly homotopic to the wedge product of $n$-circles.
Best Answer
This comes from the following theorem:
Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points. Given any point $(x_0, y_0) \in X \times Y$, the space $X \times Y -\{(x_0,y_0)\}$ is still path connected.
In your case, $X \times Y$ must be path connected since $\mathbb{R}$ is. Hence, each of $X$ and $Y$ are path connected, so the above result is applicable. But upon removing a point we get into trouble, since $\mathbb{R}-pt$ is no longer path connected, but $X \times Y -\{(x_0,y_0)\}$ is. So all you have to do is believe in the theorem.
Sketch Proof. Let $(a,b)$ and $(c,d)$ be any two points in $Z=X \times Y -\{(x_0,y_0)\}$. We must show there exists a path between these two points, the path living in $X \times Y -\{(x_0,y_0)\}$ (so, avoiding $(x_0,y_0)$). The argument comes in several cases.
First, we could have $a=c\neq x_0$. Use the fact that $\{a\} \times Y\cong Y$ is path connected and build the path in that slice. That clearly gives a path in $Z$. The same argument works if $b=d\neq y_0$.
Next, the case $a=c=x_0$. It follows that $b \neq y_0$ and $d \neq y_0$. Here's where you need the assumptions that $|X|, |Y| \geq 2$. We can choose a point $x'\in X$ which is different from $x_0$. Now create the needed path in three segments:
$$ (a,b) \to (x',b) \to (x',d) \to (c,d) $$ working in the obvious coordinate slices in each leg (see the argument for the first case). This builds a path $(a,b) \to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.
Finally, there is the case $a \neq c$ and $b \neq d$. Construct two different three-legged paths $$ (a,b) \to (c,b) \to (c,d) $$ and $$ (a,b) \to (a,d) \to (c,d). $$ It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done.