Are the plane and the third dimensional space homeomorphic? I think they are homeomorphic, intuitively.
But how can I show this intuitively, with words?
Or am I wrong? Then why?
 A: No, the fundamental group of plane-point is $Z$ and the fundamental group of $R^3$-point is $\{1\}$.
You can also say that if you remove a circle to $R^2$, the space you obtain is not connected. But if you remove a circle to $R^3$ you still have a connected space.
A: https://en.wikipedia.org/wiki/Planar_graph 
We can paint K5 in  R3, but not in R2
A: They are not homeomorphic. Deleting a straight line from $\mathbb R^2$ leaves it disconnected but that doesn't disturb the connectedness in $\mathbb R^3$.
A: — No, they are not homeomorphic (for the usual topologies). Suppose that $h : \Bbb R^2 \to \Bbb R^3$ is a homeomorphism.
Then $$h' : X=\Bbb R^2 \setminus \{(0,0)\} \to Y=\Bbb R^3 \setminus \{h(0,0)\}$$ would be also a homeomorphism.
[Indeed, it is well-defined: if $p \neq (0,0)$ then $h(p) \neq h(0,0)$ (because $h$ is injective), so that the image by $h$ of any point of $X$ lies in $Y$. As a restriction of a continuous map, $h'$ is also continuous. It is also injective, because $h$ is injective. Moreover, it is surjective onto $Y$ because $h$ is surjecive. Finally the inverse of $h'$ is continuous because it is the restriction to $Y$ of the inverse $h^{-1}$ of $h$, which is continuous (because $h$ is a homeomorphism).]
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— However, $Y$ is simply connected, while $X$ is not. In $X$, the loop $$\gamma : t \mapsto (\cos(2\pi t), \sin(2\pi t))$$ can't be deformed continuously to get the constant path $c : t \mapsto (1,0)$. 
This idea is that if you want to "shrink" the circle (which is the image of $\gamma$) into a point, you would have to pass through the origin $(0,0)$... which doesn't belong to our space $X$ !
Another way would be to "cut" the circle... but then it is not a continuous deformation anymore.
— On the other hand, any path in $Y$ can be deformed continuously into a constant loop (or into any other path).
This argument shows that $X$ can't be homeomorphic to $Y$ (because simply connectedness is preserved under homeomorphisms).

In order to prove in details that $X$ is not simply connected, I think that you need to compute its fundamental group, which happens to be $\Bbb Z$. More precisely, $$\pi_1(X) = \{\underbrace{\gamma * \cdots * \gamma}_{n \text{ times}} \;|\; n \in \Bbb Z\},$$ where $*$ denotes the concatenation of paths ;
if $n<0$ we agree that we concatenate the opposite of $\gamma$.

You can actually show that $\Bbb R^n$ is homeomorphic to $\Bbb R^m$ if and only if $n=m$, but this is not obvious. You can do it using singular homology, for instance. (In fact, using homology you can prove that any non-empty open set of $\Bbb R^n$ is not homeomorphic to any non-empty open set of $\Bbb R^m$, unless $n=m$).
