I want to find two closed, non-homeomorphic subsets $A$ and $B$ of $\mathbb{R}$ (with subset topology), with the property that there exist two continuous bijections $$f:A\to B,~~~~g:B\to A.$$
Clearly $A$ or $B$ cannot be bounded. But I didn't find more restrictions. Do we have some results on this question?
Using a relatively recent theorem, we can get a whole family of examples.
Suppose $A$ and $B$ are two closed subsets of the real line satisfying the following properties:
zero-dimensional
$\sigma$-compact, but not compact
no isolated points
Then they are "bijectively related" in the sense of your question. A proof can be found in Section 3 of this paper: https://wrbrian.files.wordpress.com/2012/01/cumet2.pdf.
It can be shown that there are $2^{\aleph_0}$ homeomorphism types of spaces satisfying these conditions (though, sadly, I don't have a reference). So now you have uncountably many examples!
An example, I hope.
Write $X \sqcup Y $ for disjoint union, say a set made up of two disjoint closed parts, one homeomorphic to $X$ and one homeomorphic to $Y$. And write $\bigsqcup_{i} X_i$ for a disjoint union of countably many closed sets homeomorphic to the $X_i$, "going to infinity" in the sense that any bouned interval meets only finitely many of them. Write $X \prec Y$ to mean there is a continuous bijection from $X$ to $Y$.
Let $P$ (or with subscripts) be a single point.
Let $Q$ (or with subscripts) be homeomorphic to $\{0\}\cup\{3^{-n}:n=1,2,3,\cdots\}$; a convergent sequence together with its limit.
Let $C$ (or with subscripts) be homeomorphic to the middle-thirds Cantor set in $[0,1]$.
My two sets are $$ A = \bigsqcup_i P_i \sqcup \bigsqcup_i C_i \\ B = \bigsqcup_i P_i \sqcup \bigsqcup_i C_i \sqcup Q $$ Of course $A$ and $B$ are homeomprphic to closed subsets of $\mathbb R$. For example, $A$ could consist of the negative integers, together with Cantor sets in each of the intervals $[2i,2i+1]$.
The derived set $A'$ of $A$ is $\bigsqcup_i C_i$, and has no isolated point. But the derived set $B'$ of $B$ is $\bigsqcup_i C_i$ plus a single isolated point $Q'$. So $A$ and $B$ are not homeomprphic.
Now we have to observe $$ \bigsqcup_i X_i = \bigsqcup_i X_{2i} \sqcup \bigsqcup_i X_{2i+1} \tag{0} $$
$$ \bigsqcup_i P_i \prec Q \tag{1} $$
$$ \bigsqcup_i C_i \sqcup P \prec C \tag{2} $$
$$ \bigsqcup_{i} C_{i} \sqcup Q \prec C \tag{3} $$ From these we get $A \prec B$ and $B \prec A$: $$ A = \bigsqcup_i P_i \sqcup \bigsqcup C_i = \bigsqcup_i P_i \sqcup\bigsqcup_i P_i \sqcup \bigsqcup C_i \prec Q \sqcup\bigsqcup_i P_i \sqcup \bigsqcup C_i = B $$
$$ B = \bigsqcup_i P_i \sqcup \bigsqcup C_i \sqcup Q = \bigsqcup_i P_i \sqcup \bigsqcup C_i\sqcup \bigsqcup C_i \sqcup Q \\ \prec\bigsqcup_i P_i \sqcup \bigsqcup C_i\sqcup C =\bigsqcup_i P_i \sqcup \bigsqcup C_i = A $$
