Continuous bijection from $\mathbb R-\{0\}$ to $\mathbb R$ I know such a function exists if continuity is not required.

How do you prove it doesn't exist when continuity is required? 

For the sake of completeness,  note that a continuous bijection from $\mathbb R$ to $\mathbb R-{0}$ cannot exist either (the IVT applies trivially).
Assume $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ is a continuous bijection. Denote $\mathbb{R}_{&gt;0} = \{x \in \mathbb{R} \mid x &gt; 0\}$ and $\mathbb{R}_{&lt;0} = \{x \in \mathbb{R} \mid x &lt; 0\}$.


Spaces $\mathbb{R}_{&gt;0}$ and $\mathbb{R}_{&lt;0}$ are connected and $f$ is continuous, therefore $f(\mathbb{R}_{&gt;0})$ and $f(\mathbb{R}_{&gt;0})$ are also connected. Since $f$ is a bijection, we see that $\mathbb{R}$ is a union of two disjoint non-empty connected subsets: $\mathbb{R} = f(\mathbb{R}_{&lt;0}) \cup f(\mathbb{R}_{&gt;0})$.
Connected subsets in $\mathbb{R}$ are precisely the same thing as intervals. It's easy to see then that one of the two sets $f(\mathbb{R}_{&lt;0})$ and $f(\mathbb{R}_{&gt;0})$ must be unbounded on one side and closed on the other side. For instance, let's assume that $f(\mathbb{R}_{&gt;0}) = [y, +\infty)$, where $y \in \mathbb{R}$. (The other possible cases are analogous).
Let $x = f^{-1}(y)$. The restricted function $f|_{\mathbb{R}_{&gt;0}}$ reaches its minimum at point $x &gt; 0$. It follows that there are $x_1$ and $x_2$ somewhere close to $x$ such that $0 &lt; x_1 &lt; x &lt; x_2$, $f(x_1) &gt; f(x)$ and $f(x_2) &gt; f(x)$. Now it's easy to reach a contradiction with the injectivity of $f$ using the intermediate value theorem. Done.
The topology is different. $\mathbb{R}=(-\infty,\infty)$ has two open ends and $\mathbb{R}-\{0\}=(-\infty,0)\cup(0,\infty)$ has four. Continuous functions map boundaries to boundaries, which cannot happen bijectively. For instance, $\log |x|$ is an example that maps two-to-one.

Continuous bijection only exists between spaces with equal topology.