Continuous bijective function between the same topology that is not a homeomorphism. I know there are many examples when the domain and co-domain do not coincide. Taking the identity on $X$ from $(X,\tau_1)$ to $(X,\tau_2)$ when $\tau_2$ is coarser than $\tau_1$ gives an infinite family of examples.
However I have been struggling to find an example of such a function between the same topology. I came to think about this because in $\mathbb R$ any bijective continuous function is a homeomorphism.
Thank you in advance, regards.
 A: The simplest - in the sense of least things to verify - examples that I can think of are linear operators on some infinite-dimensional vector spaces. For example, endow the space
$$c_{00}(\mathbb{N}) := \bigl\{ x \colon \mathbb{N}\to \mathbb{C} : \bigl(\exists k\in\mathbb{N}\bigr)(n \geqslant k \implies x(n) = 0)\bigr\}$$
with an $\ell^p$-norm for $1 \leqslant p \leqslant +\infty$, and consider the operator $T$ given by
$$T(x)(n) = 2^{-n}\cdot x(n).$$
Then $T$ is linear, continuous and bijective (easily checked), but its inverse is unbounded, hence not continuous.
A: Let $\tau$ be the usual topology of $\mathbb R.$ Define a finer topology $\tau_1$ as follows:
$$\tau_1=\{X\subseteq\mathbb R:X\cap(0,\infty)\in\tau\}.$$
The function $f(x)=x+1$ is a continuous bijection, but not a homeomorphism, from $(\mathbb R,\tau_1)$ to itself.
A: Let $K=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the topology that it inherits from $\Bbb R$, let $\Bbb N$ have the discrete topology, and let $X=\Bbb N\times K$ with the product topology. Let 
$$A=\{\langle n,y\rangle\in X:n\text{ is odd}\}\;;$$
$A$ is countably infinite, so let $\varphi:\Bbb N\times\{1\}\to A$ be any bijection. Then
$$f:X\to X:\langle n,y\rangle\mapsto\begin{cases}
\langle 2n,y\rangle,&\text{if }y=0\\
\left\langle 2n,\frac1{k-1}\right\rangle,&\text{if }y=\frac1k<1\\
\varphi(\langle n,1\rangle),&\text{if }y=1
\end{cases}$$
is a continuous bijection that is not open: $f$ maps the isolated point $\varphi^{-1}(\langle 0,0\rangle)$ to the non-isolated point $\langle 0,0\rangle$.
This is simpler than it may at first appear. Pictorially, $X$ is just the disjoint union of countably infinitely many simple sequences:
$$\begin{array}{c|cc}
0&\color{brown}\bullet&\color{brown}\bullet&\color{brown}\bullet&\color{brown}\bullet&\ldots\\
\uparrow&\uparrow&\uparrow&\uparrow&\uparrow&\cdots\\
\frac14&\bullet&\bullet&\bullet&\bullet&\cdots\\
\frac13&\bullet&\bullet&\bullet&\bullet&\cdots\\
\frac12&\bullet&\bullet&\bullet&\bullet&\cdots\\
1&\bullet&\bullet&\bullet&\bullet&\cdots\\ \hline
&0&1&2&3&\cdots
\end{array}$$
$A$ consists of the odd-numbered columns. The map $f$ takes all of column $n$ except the bottom point to column $2n$, moving the isolated (black) points down one but leaving the limit (brown) points at the top; the (blue) point at the bottom of column $1$ goes somewhere in $A$. 
$$\begin{array}{c|cc}
0&\color{brown}\bullet&\rightarrow&\color{brown}\bullet\\
&\uparrow&&\uparrow&\\
\frac14&\bullet&&\bullet\\
&&\searrow\\
\frac13&\bullet&&\bullet\\
&&\searrow\\
\frac12&\bullet&&\bullet\\
&&\searrow\\
1&\color{blue}\bullet&&\bullet\\ \hline
&1&&2
\end{array}$$
This maps all of $X$ except the bottom row bijectively (and homeomorphically) onto $X\setminus A$, the even columns. The points in the bottom row are isolated, so a continuous map can send them anywhere; $\varphi$ (and hence $f$) simply sends them bijectively to $A$.
