Easy example of a bijective continuous self mapping whose inverse is discontinuous Let $f : X \to X$ be a continuous bijective mapping from a metric space onto itself.
Is $f^{-1}$ continuous too? I don't think so, but I'm struggling to find a counterexample.
I've read that if $X=\mathbb{R}^n$ or $X$ a compact Hausdorff space, $f^{-1}$ is always continuous. The Open Mapping Theorem gives us another restriction to find a counterexample, namely all linear operators from a Banach space to itself.
Also, I know there's an easy example if we allow to use two different topologies. Here I'm only considering the topology induced by the one metric we have.
 A: Consider the space$$c_{00}=\left\{(x_n)_{n\in\Bbb N}\,\middle|\,\bigl((\forall n\in\Bbb N):x_n\in\Bbb C\bigr)\text{ and }x_n=0\text{ if }n\gg0\right\},$$endowed with the $\sup$ norm, and define $f\colon c_{00}\longrightarrow c_{00}$ by$$f\bigl((x_n)_{n\in\Bbb N}\bigr)=\left(\frac{x_n}n\right)_{n\in\Bbb N}.$$Then $f$ is bijective and continuous, but $f^{-1}$ is discontinuous. Actually, it is discontinous everywhere.
A: There are already some nice answers about linear maps. I'd like give you an example which is not linear. The main idea is if you have a bijective map such that $f((0,1]\cup (2,3])= (0,1]$, then the inverse cannot be continuous as it needs to tear the interval $(0,1]$ apart (not possible as continuous functions map connected sets to connected sets).
The counterexample with an incomplete metric space: The standard example for a continuous map where the inverse is not continuous would be something like this:
$$ f: (0,1]\cup (2,3] \rightarrow (0,1]\cup (2,3], \ f(x)=\begin{cases} f(x)=\frac{1}{2} x,& x\in (0,1], \\ \frac{1}{2} x - \frac{1}{2},& x\in (2,3].
\end{cases} $$
The map is injective as it is strictly monotone increasing. Furthermore it is clearly continuous (it is local a polynomial function). However, its inverse is not continuous as $f((0,1])= (0,1] \cup (2,3]$ (or you could explicitely compute the inverse and see that it has a jump at $1/2$).
Of course the example is not surjective. Let's fix this by slightly changing the domain. We consider
$$ X:= \bigcup_{n\in \mathbb{N}} (2n, 2n+1].$$
Now all we have to do is extending our previous function to this new domain. We will call the extension $g: X\rightarrow X$.
For $x\in (0,1]\cup (2,3]$ we define $g(x)=f(x)$. Furthermore, we map $(2n+4, 2n+5]$ continuously and bijectively to $(2n+2, 2n+3]$ (for example by just shifting, i.e. $g(x)= x-2$), then we obtain a function as you desire. Indeed, it is a bijection from $(0,1]\cup (2,3]$ to $(0,1]$ and bijection from $\bigcup_{n\in \mathbb{N}_{\geq 2}} (2n, 2n+1]$ to $\bigcup_{n\in \mathbb{N}_{\geq 1}} (2n, 2n+1]$ and hence $g$ is a bijection on all of $X$. The function is locally a polynomial function and hence continuous. Finally the inverse function is not continuous by the same argument that the inverse of $f$ is not continuous.
Counterexample for complete metric space (added afterwards): In the comments @KaviRamaMurthy asked whether one could construct a counterexample where the metric space is complete. Indeed one can. We use a similar idea as above. We also want to glue together connected components. We cannot use half-open intervals in $\mathbb{R}$, however, we can use two halflines and one point in $\mathbb{R}^2$. Namely, we take
$$ X:= \bigcup_{n\in \mathbb{N}} \big((-\infty, -2]\times \{ n \} \cup [2,\infty) \times \{ n \} \big) \cup \bigcup_{n\in \mathbb{N}} \{ (0,n) \}
\cup \big([-1,1] \times \{ -1 \} \big) \cup \bigcup_{n\in \mathbb{N}_{\geq 2}} \big([-1,1] \times \{ -n \} \big)\subseteq \mathbb{R}^2. $$
As $X$ is a closed subset of $\mathbb{R}^2$, we get that it is a complete metric space. We map the two lowest halflines $\big((-\infty, -2]\cup [2, \infty)\big)\times \{0\}$ and the lowest point $(0,0)$ bijectively and continuously to the interval $[-1,1]\times \{-1\}$ (use $f(x,0):= \left(\frac{\operatorname{sgn}(x)}{\vert x \vert-1}, -1 \right)$ for the halflines and $f(0,0):=(0,-1)$). Then we shift the remaining levels down, i.e. we just take $f(x,n+1) :=(x, n)$ for all $(x,n+1)\in X$ with $n\in \mathbb{N}$. Finally we shift also all the compact interval down, i.e. $f(x,-n):=(x,-n-1)$ for all $(x,-n)\in X$ and $n\in \mathbb{N}_{\geq 2}$.
This yields a continuous, bijective function $f: X \rightarrow X$. However, its inverse is not continuous as it "tears" the interval $[-1,1]$ into three pieces. The easiest is to quickly draw the corresponding picture (I would do it here, but I am terrible in drawing things on the computer). As above, the real thing happens when we glue the two halflines and the point together to the interval. The remaining "levels" are just there to make the map a bijection.
The counterexample in any infinite-dimensional normed space (added even later): 
First we note that we cannot have any such counterexample in any finite-dimensional normed space. Indeed, they are all homeomorphic to some $\mathbb{R}^n$ and the invariance of the domain theorem (https://en.wikipedia.org/wiki/Invariance_of_domain) tells us that any continuous, injective map $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a homeomorphism on its image.
In 
Van Mill, J. (1987). Domain Invariance in Infinite-Dimensional Linear Spaces. Proceedings of the American Mathematical Society, 101(1), 173-180. doi:10.2307/2046571
it is claimed that we can construct such an example in any infinite-dimensional normed space $X$. The only thing one needs to construct is a continuous map $\lambda : S \rightarrow (0,1]$ with $\inf_{x\in S} \lambda(x) =0$, where $S$ is the unit sphere in $X$. Then one considers the map 
$$f: X \rightarrow X, \ f(x) = \begin{cases} \lambda\left( \frac{x}{\Vert x \Vert} \right) x,& x\neq 0, \\
0,& x=0.
\end{cases}$$
The function $f$ is clearly bijective and continuous (as $\lambda$ is continuous and does not vanish). To see that it is not a homeomorphism we show that it is not an open map. This follows from the fact that $f$ preserves all halfrays starting in zero and that by construction that we find halfrays that shrink as much as we like (as $\inf_{x\in S} \lambda(x) =0$). So for any bounded set $A\subseteq X$ we have $0\notin \operatorname{int}(f(A))$ and thus $f$ is not an open map and hence not a homeomorphism.
We are left to show that such a $\lambda$ really exists. As we are in infinite-dimensions, by the Riesz lemma (https://en.wikipedia.org/wiki/Riesz%27s_lemma) there exists a sequence $(x_n)_{n\in \mathbb{N}} \subseteq S$ such that $\Vert x_n - x_m \Vert \geq \frac{1}{2}$ for all $n\neq m$. Now we make them slightly bigger, i.e. we define $y_n:= \left( 1+ \frac{1}{n+4} \right) x_n$. Then we have that $dist(S, y_n)=1/(n+1)$ and for $n\neq m$
$$\Vert y_n - y_m \Vert \geq \Vert x_n - x_m \Vert - \frac{1}{n+4} - \frac{1}{m+1} \geq \frac{1}{2} - \frac{1}{4} - \frac{1}{5} = \frac{1}{20}. $$
Hence, $Y:=\{ y_n \ : n\in \mathbb{N} \}$ is discrete and hence closed. Now we define (inspired by the proof of the Urysohn lemma)
$$ g: S \rightarrow [0,\infty), \ g(x) := dist(x,Y). $$
Hence, $g$ is continuous and $g(x) = 0$ iff $x\in \overline{Y}=Y$. This shows that $g$ is continuous and never vanishes. Furthermore, we have 
$$g(x) = dist(x,Y) \leq \Vert x - y_0 \Vert \leq \Vert x \Vert + \Vert y_0 \Vert  
\leq 1 + \left( 1 + \frac{1}{4}\right) = \frac{9}{4}.  $$
Thus, we may pick $\lambda = \frac{4}{9} g$. 
Added in the end: One could also construct an example with $X\subseteq \mathbb{R}$ complete metric space. This is done here Example of a bijective continuous self mapping whose inverse is not continuous on a complete subspace of $\mathbb{R}$
A: A standard example of this is the map $f \to \int_0^{x} f(t)dt$ on the space $C[0,1]$ with the supremum norm. Note that the image if $x^{n}$ converges in this space but $x^{n}$ itself does not converge, Hence the inverse is not continuous.
Another example: Let $X=\ell^{2}$ and $f((x_n))=((\frac 1 n x_n))$. Then $f$ is continuous and injective, $f(e_n) \to 0$ but $e_n$ does not tend to $0$. Here $e_1,e_2,...$ is the standard basis of $\ell^{2}$. 
