is $\mathbb{R}^2\setminus \{(0,0)\}$ homeomorphic to $S^1$?

Removing $2$ points will disconnect $S^1$, but not $\mathbb{R}^2 \setminus \{(0,0)\}$. There is, however, a retraction $r : \mathbb{R}^2 \setminus \{(0,0)\} \to S^1$ given by $x \mapsto \frac{x}{\|x\|}$.

They are not homeomorphic.

Suppose that they are, and let $f:\Bbb R\setminus\{(0,0)\}\to \Bbb S^1$ an homeomorphism.

Let $x=(1,0)$ and $y=(0,1)$. It is clear that $Y=\Bbb S^1\setminus\{x,y\}$ is not connected, but $f^{-1}(Y)$ is the plane with three points removed, that is connected.

$\mathbb{S}^1$ is compact, continuous image of a compact set is compact and $\mathbb{R}^2/(0,0)$ is not compact.

So you cannot find a continuous surjection $f:\mathbb{S}^1 \rightarrow \mathbb{R}^2/(0,0) $

No, they are not homeomorphic, since they have different dimension. Although they are homotopically equivalent: you can see that $S^{1}$ is a strong deformation retract of $\mathbb{R}^2\setminus(0,0)$.