Proving 2-sphere is not homeomorphic to plane

I need to prove 2-sphere is not homeomorphic to plane. I tried to use connectedness. $S^2$ and $R^2$ are both connected. If they were homeomorphic when we throw out same points from this space then they still must be homeomorphic and connected to. so where $f\colon S^2 \to R^2$, is bijection and $f$ is bicontinuous. But I am not sure $S^2\setminus \{(0,0)\}$ and $R^2\setminus \{(f(0),f(0))\}$ are connected or not. Could you help at this point?

As @carmichael561 noted, it is much easier to just use the fact that $S^{2}$ is compact. Thus, if there were a continuous bijection $f\colon S^{2}\to \mathbb{R}^{2}$, then $\mathbb{R}^{2}=f(S^{2})$ would have to be compact, since the image of a compact set under a continuous function is compact. Since $\mathbb{R}^{2}$ is not compact, no such $f$ can exist.

The set of open disks of radius $1$ in the plane is an open cover. Every finite subset of it is bounded and therefore cannot cover the whole plane. Hence the plane is not compact.

The $2$-sphere is the inverse image of the closed set $\{1\}$ under the continuous function $(x,y,z)\mapsto x^2+y^2+z^2$ and is therefore a closed subset of $\mathbb R^3.$ It is also bounded. Being a closed and bounded subste of $\mathbb R^3,$ it is compact.

Perhaps one should add that compactness is a topological property: any space that is homeomorphic to a compact space is compact.

Since many good answers were already given I try giving a bit of a conceptual one as well as some interesting answers (in the sense that they aren't the standard ones).

Consider any finite dimensional real vector space $\mathbb R^n$, with $n \in \mathbb N_{\geq 1}$, and equip it with the standard topology induced by any metric on $\mathbb R^n$. This space is hausdorff, locally compact but not compact. There's a straight ahead proof of this statement. This suffices to show that $\mathbb R^n \not\cong S^n$. If you are familiar with the stereographic projection you know that $\mathbb R^n$ is homeomorphic to the punctured sphere $S^n \setminus\{x\}$ with $x \in S^n$. Now consider $f:S^n \rightarrow \mathbb R \cup \{\infty\}$ such that $f|_{S^n\setminus\{x\}}:S^n \setminus\{x\}\rightarrow \mathbb R^n$ is a homeomorphism and $f(x)=\infty$. We now choose the topology $\mathcal O_{R \cup \{\infty\} }:=\{O\in P(\mathbb R \cup \{\infty\} ) \mid \exists \check o \in \mathcal O_{S^n}: f(\check o)=O\}$. This makes the bijective map $f$ a homeomorphism and thus $\mathbb R^n \cup \{\infty\}$ a compact space. From the fact that $f|_{S^n\setminus\{x\}}:S^n \setminus\{x\}\rightarrow \mathbb R^n$ was choosen to be a homeomorphism it follows that the subspace $\mathbb R^n\subset \mathbb R^n\cup\{\infty\}$ equipped with the subspace topology is homeomorphic to $\mathbb R^n$. Thus we can think of the $n$-sphere as the one-point compactification of $\mathbb R^n$. we can consider its one-point compactification $\mathbb R^n \cup \{\infty\}$.

Now for a more "fun proof" : Consider any continous mapping $f:S^n \rightarrow \mathbb R^n$. The Borsuk-Ulam theorem states that there is an $x \in S^n$ such that $f(x)=f(-x)$. Be aware that $x$ and $-x$ are well defined and different from each other. This already tells you that there can't be an injective continous map from the $n$-sphere to $\mathbb R^n$. The trick here is obviously understanding the Borsuk-Ulam theorem and may serve as some motivation for algebraic topology.

Another rather unusual proof is with methods from homotopy theory. Assume $\mathbb R^2$ and $S^2$ to be homeomorphic and $f: \mathbb R^2 \rightarrow S^2$ beeing such a homeomorphism. Now consider the subspace $\mathbb R^2\setminus\{0\}$ with the subspace topology. If $f$ is a homeomorphism $\mathbb R^2\setminus\{0\}$ must be homeomorphic to $S^2\setminus \{f(0)\}$ when both spaces are equipped with the subspace topology. If we consider any closed path $\phi:[0,1] \rightarrow \mathbb R^2 \setminus\{0\}$ encirceling $0$ we cannot contract this path to a constant one. However the path $f\circ \phi: [0,1]\rightarrow S^2\setminus\{f(0)\} $ encircles $f(0)$ but can be contracted. This is a contradiction to the hypothesis of $f$ beeing a homeomorphism.

There is a "sort of" connectedness argument, except the obvious compactness argument. Leaving out points is one such idea:

By the stereographic projection we know that $\mathbb{S}^2 \setminus \{(0,0,1)\}$ is homeomorphic to $\mathbb{R}^2$, which is contractible and so simply connected. On the other hand, $\mathbb{R}^2 \setminus \{p\}$ is not simply connected, as it has the circle $\mathbb{S}^1$ as a deformation retract and the circle is not simply connected. Otherwise put, if $h: \mathbb{S}^2 \to \mathbb{R}^2$ were a homeomorphism then $\pi_1(\mathbb{S}^2 \setminus \{p\}) \simeq \pi_1(\mathbb{R}^2 \setminus \{h(p)\})$, for any $p \in \mathbb{S}^2$, but $\pi_1(\mathbb{S}^2 \setminus \{p\}) \simeq \{0\}$ while $\pi_1(\mathbb{R}^2 \setminus \{p\}) \simeq \mathbb{Z}$, giving us a contradiction.

So with connected replaced by "simply connected" your idea will work. But this involves much more advanced notions, so it's not that "simple", despite the name. One could see a lot of algebraic topology as an extension of such ideas. The homotopy groups are a sort of quantification of a higher order connectedness, you could say.