Why $\Bbb{R}$ and $\Bbb{R}^2$ are not homeomorphic. Can I argue that if there exists a homeomorphism
$$f: \Bbb{R} \rightarrow \Bbb{R}^2$$
Then subtracting a point should preserve connectedness by continuity of $f$, but then $\Bbb{R}$ minus the origin is disconnected while $\Bbb{R}^2$ minus the origin Is still connected. Is this a good enough argument? As connectedness is a topological property. Which I can prove.
 A: I agree with the commenters that your argument is basically right, however there is one subtlety:
We don't know that $f(0) = 0$. But when we remove a point in $\mathbb{R}$ we should make sure we're removing the corresponding point (as given by $f$) in $\mathbb{R}^2$.
So a slightly better argument would be to say that $\mathbb{R} \setminus \{ 0 \}$ is disconnected, whereas $\mathbb{R}^2 \setminus \{ f(0) \}$ is connected.
Obviously, though, your idea still works.

I hope this helps ^_^
A: Let $f : \mathbb{R}^2 → \mathbb{R}$ is a homeomorphism. Then, restricting the domain to $\mathbb{R}^2−\{(0,0)\}$ gives a homeomorphism of the punctured euclidean space to $\mathbb{R}− {f(\{(0,0)\})}$. However, the punctured euclidean space is connected, but as $\mathbb{R}− {f(\{(0,0)\})}$ is not connected, because note that
$$\mathbb{R}− {f(\{(0,0)\})} = (−\infty, f(\{(0,0)\})) \cup (f(\{(0,0)\}),\infty)$$
so the open sets $(−\infty, f(\{(0,0)\}))$ and $(f(\{(0,0)\}),\infty)$ give a separation of this space.
Hence, the punctured euclidean space and  $\mathbb{R}− {f(\{(0,0)\})}$ are not homeomorphic, a contradiction. Therefore, we conclude that $\mathbb{R}^2$ and $\mathbb{R}$ are not homeomorphic.
