For some classical spaces in topology there has been a long research tradition of finding nice unique characterisations of a space, i.e. a finite list of "real" properties (directly defined in terms of open sets and set theory, like compactness and connectedness (and their local versions) and various ones) such that a space that satisfies that finite list of properties is homeomorphic to the space in question. To be useful, it should be non-circular ones (so being a simple-closed curve is not a "good" property to characterise the circle, because that's already defined as being homeomorphic to a circle, basically).
For zero-dimensional spaces we have the classic ones for the Cantor set (the unique zero-dimensional compact metrisable space without isolated points), the rationals (the only countable metrisable space without isolated points), the irrationals (the only completely metrisable separable zero-dimensional, nowhere locally compact space) and a few more.
For connected metric spaces (often continua) we have classic characterisations of $[0,1]$ (metrisable Peano-continuum with exactly two non-cutpoints), $\mathbb{R}$ (separable metrisable, connected, locally connected and every point a strong cut-point (exactly two components of the space minus that point)), $S^1$( metrisable Peano continuum, with no cutpoints and such that every pair is a cutset) and a few others.
For infinite-dimensional spaces we have the topological classification of completely metrisable separable vector spaces (they're all homeomorphic, so purely topologically there is no difference between $C([0,1])$ and separable Hilbert space, e.g.)
But for general spaces there is no hope of such results, they are just too wild and have lots of connections with complicated set theory. By transfinite recursion one can sometimes construct non-homeomorphic spaces of very similar type (so that they'd be the same for a list of standard properties, at least that I could think of (I did this kind of thing in my PhD thesis) but that nonetheless are non-homeomorphic because they are constructed that way: by a diaginalisation argument one can often enumerate in advance all possible candidate functions, and then during the construction ensure that none of them will be a homeomorphism of the eventually resulting space; it's a common trick topologists use, Sierpinski already did it back in the 1920's). So sometimes spaces are not homeomorpjic because they aren't by construction. And if I made $2^{\aleph_1}$ distinct spaces in some construction, there is no way I could distinguish them even using a finite list from $\aleph_1$ properties, if could invent that many "properties" in the first place.. The only option left would be to give up and allow "homeomorphic to $X$" as a (useless) property for all $X$.
Just my thoughts..
Copying part of my answer to this Math Overflow question:
Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.
Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):
"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."
Best Answer
Well, it usually goes the other way. A property $P$ of a topological space $X$ is deserved to be called topological if $P(X)$ holds if and only if $P(Y)$ holds whenever $X$ and $Y$ are homeomorphic. An example of a topological property is "$X$ is connected" while an example of a non-topological property is "$X$ is a subset of $\mathbb{R}^n$. The latter can be upgraded to a topological property by requiring instead "$X$ can be embedded in $\mathbb{R}^n$".
With this convention, given a topological space $X$ there is a smart-ass topological property one can define using $X$: "$P(Z)$ holds iff $Z$ is homeomorphic to $X$". Since homeomorphism is an equivalence relation, this is indeed a topological property and clearly $X$ satisfies $P$. If $Y$ is another space that satisfies the same topological properties as $X$, then $P(Y)$ must hold and so $X$ and $Y$ are homeomorphic.