Let $(X,\tau)$ be a topological space. If $A\subseteq X$ we define the following equivalence relation on $X$: $$\sim\_A = \{(x,y) \in X^2: x=y \text{ or }\{x,y\}\subseteq A \}.$$


Let $(X,\tau)$ be an infinite space with the following property:



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> If $A\subseteq X$ and $|A|<|X|$ then $X\cong X/\sim\_A$.
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> 
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Does this imply that $(X,\tau)$ is one of the following?


1. $(X,\tau)$ is discrete;
2. $(X,\tau)$ is indiscrete;
3. $X$ can be well-ordered with a relation $\leq$ such that $\tau$ consists of the [upper sets](https://en.wikipedia.org/wiki/Upper_set) of $(X,\leq)$.



Spaces that are invariant under some contractions
For a countable space $X$ the discussed property means that $X$ is homeomorphic to $X/A$ for any finite subset $A\subset X$. Let us call this property *quotient-homogeneous*. 


There are many quotient-homogeneous countable spaces which do not fall into the categories (1)-(3) of Dominic van der Zypen:


4) any countable $T\_1$-space with a unique non-isolated pont;


5) the space of rational numbers.


It seems that many (but not all) countable ordinals endowed with the interval topology are quotient-homogeneous.


Observe that spaces of form (3) do not satisfy the separation axiom $T\_1$. Maybe Dominic van der Zypen had in mind the interval topology (not the upper-set topology)? But anyway the space of rationals is a counterexample. 


The space pointed by @Keith Kearnes is another (non-metrizable) counterexample.


