TITLE: Non-homeomorphic spaces such that taking away a point makes them homeomorphic
QUESTION [19 upvotes]: Are there topological spaces $X,Y$, each having more than $2$ points, satisfying the following two properties?

$X\not\cong Y$, and
there is a bijection $\varphi: X\to Y$ such that for all $x\in X$ the spaces $X\setminus \{x\}$ and $Y\setminus \{\varphi(x)\}$ are homeomorphic.

REPLY [2 votes]: For a minimal example consider the following two 3-element posets equipped with the Alexandroff topology (i.e. open sets are down-sets):
(a)   (b)                      (C)
  \   /                        / \
   \ /                        /   \
   (c)                      (A)   (B)

The bijection is given by: $a\mapsto A$, $b\mapsto B$, $c\mapsto C$.
The reconstruction conjecture for ordered sets (see e.g. Jean-Xavier Rampon, What is reconstruction for ordered sets?) - together with equivalence of finite posets and finite $T_0$ topological spaces (via the  specialization order/Alexandroff topology functors) - says the are no such examples among finite $T_0$ spaces with at least four elements. However, the conjecture is open.
(The related reconstruction conjecture for graphs is better known and also open.)