Recycling an old (ca. 1998) sci.math post:
" Anyone know an example of two topological spaces $X$ and $Y$ with continuous bijections $f:X\to Y$ and $g:Y\to X$ such that $f$ and $g$ are not homeomorphisms?
Let $X = Y = Z \times \{0,1\}$ as sets, where $Z$ is the set of integers. We declare that the following subsets of $X$ are open for each $n>0$. $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a topology on $X$.
We declare that the following subsets of $Y$ are open for each $n>0$. $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ This is a basis for a toplogy on $Y$.
Define $f:X\to Y$ and $g:Y\to X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i).$ Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.
This example is due to G. Paseman.
David Radcliffe "
More generally, take a space X with three successively finer topologies T, T' and T''. Form two spaces which have underlying set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' .... and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer topology in one sequence to a rougher topology in the other. You can make them bijective, and show that they are obviously non-homeomorphic for a judicious choice of X, T, T', and T''.
Gerhard "Ask Me About System Design" Paseman, 2010.07.05
