Proving Homeomorphisms between Topological Spaces without finding explicit functions I am starting to study General Topology, and until now I have solved exercises involving proving homeomorphisms between spaces (such as S¹ and a square) by defining a bijective function $f: X \rightarrow Y$ between the spaces and verifying that either all open sets of $Y$ are the image of an open set of $X$ or that both $f$ and $f^{-1}$ are continuous.
However, I find that sometimes defining an explicit function that maps one space to another might be too laborious or, in fact, unnecessary. I feel that there surely are better ways to proving such things using only other known facts about the structure of the spaces. (i.e. Showing that the open interval $(0,1)$ is homeomorphic to $S¹ - \{(0,1)\}$ (the unit circle with the removal of one point) intuitively doesn't seem to require an explicit function. Maybe the fact that both have two "ends", such as any open interval on the real line should be enough?)
What I am looking for are tools or some concepts that may help me proving these types of things, and maybe the existance of bijective functions between sets in general, not only in problems involving Topological Spaces.
Thanks in advance.
 A: For "the existence of bijective functions between sets in general" (i.e., forgetting about topological structure) there are theorems like the Schroeder-Bernstein theorem that can make life easier.
For proving that two topological spaces are homeomorphic, you generally have to exhibit the homeomorphism. Fortunately, you can often use geometrical thinking as well as composition with homeomorphisms you've "prepared earlier" (as they say in cookery shows) to help with that: e.g., in your example, the 1-dimensional version of the stereographic projection gives you a homeomorphism of $S^1 \setminus \{(0, 1\}$ with $\Bbb{R}$. You can then compose this with your favourite homeomorphism of $\Bbb{R}$ with the open interval $(0, 1)$ (e.g., based on the $\mathrm{arctan}$ function) to get a homeomorphism of $S^1 \setminus \{(0, 1\}$ with $(0, 1)$.
One useful result that can save some work is that if $X$ is compact and $Y$ is Hausdorff and $f: X \to Y$ is continuous and bijective, then $f$ is a homeomorphism. (Because closed subsets of compact sets are compact, continuous functions map compact sets to compact sets and compact subsets of Hausdorff spaces are closed.)
A: There are several spaces $X$ that are so nice that there is a characterisation of them: a finite list of properties such that any space $Y$ that satisfies these properties is homeomorphic to $X$. Quite a few of these theorems were proved for classical spaces in the beginning of the field of topology, but not many afterwards. It seems that most of such spaces are very large (infinite-dimensional) or zero-dimensional (so very disconnected). In he proof of such theorems there is a construction of a homeomorphism (often by a limiting process, or recursion), but it doesn't give explicit formulas in many cases.
Some classical examples:

*

*Any countable metric space without isolated points is homeomorphic to $\Bbb Q$ in the standard topology (due to Sierpiński), a proof can be found here.


*Any compact metrisable totally disconnected space without isolated points is homeomorphic to $C$ (the middle-third Cantor set, or $\{0,1\}^{\Bbb N}$ if one prefers).


*Any connected, locally connected, separable, metric space such that every point is a strong cut-point (removing it leaves exactly two components) is homeomorphic to $\Bbb R$. (see here for references and some more info).


*Any separable completely metrisable, locally convex linear topological space is homeomorphic to $\Bbb R^{\Bbb N}$ (so this includes $\ell_2$, $C[0,1]$ any many standard spaces in analysis). Due to Anderson, and many others.
Others exists but are more technical still. But many are still open too.
As to the question of the circle and $(0,1)$: $\Bbb S^1$ has a characterisation as the unique compact connected metric space such that remving any two points leaves it disconnected. See these notes for a proof. So removing one point from $\Bbb S^1$ still leaves it separable, connected, locally connected and every point in the remainder is a strong cut-point. So it is homeomorphic to $\Bbb R$ or $(0,1)$, which are the same topologically. But this use of the theorems is overkill, as a simple projection map suffices to show it. You could also do it using a Moebius transform in the complex plane. So there it is easy to see directly. These characterisation theorems are mostly useful in the context of other proofs.
