When is a topological property useful for deciding if two spaces are homeomorphic I know that being connected is a topological property. But, my question is what are examples of two spaces where this isn't helpful?
Is this simply two spaces that are disconnected or having some other topological property? I know that that is also a topological property, but I'm asking as if connected is the only property known at this point - for the purposes of this question.
Moreover, why are so many different topological properties needed / used. Is this simply due to the wide array of topologies there can be?
 A: Here is just a bag-grab:
homology: Consider $\mathbb R^2$ and $\mathbb R^3$. These are not homeomorphic, but they are both connected, and neither have cut points.
Most of the tools for showing two objects are not homeomorphic comes from showing that they are not even homotopy equivalent. For this property, we can assign algebraic invariants, in this case homology does the trick.
cut points:
Another example is usually $S^1$ and $[0,1]$. These are connected, but if there were a homeomorphism $\phi:S^1 \to [0,1]$, then by surjectivity, there is some point $a$ that maps to $.5 \in [0,1]$, so the homeomorphism should restrict to $S^1\setminus\{a\} \to [0,1]\setminus\{.5\}$, but the latter is disconnected. This is a usual argument by cut points.
compactness: $(0,1)$ is not homeomorphic to $[0,1]$, since one is compact and one is not.
Cardinality: $\mathbb Z$ is not homeomorphic to $(n,n+1)$ with $n \in \mathbb Z$ under any topologies since they have different cardinalities (but are both really disconnected.)
There are other variations on a theme: there are linking numbers for knots, all kinds of local connectedness properties etc.
A: Similar to Andres' answer, you can show that $\mathbb R$ and $\mathbb R^2$ are not homeomorphic by the preservation of connected components.  Indeed, if there was a homeomorphism $f:\mathbb R \to \mathbb R^2$, then the restriction $f:\mathbb R\setminus\left\{0\right\} \to \mathbb R^2\setminus\left\{f(0)\right\}$ should also be a homeomorphism.  But $\mathbb R\setminus \left\{0\right\}$ is not connected, while $\mathbb R^2$ is (path) connected.
A: In order to prove that two spaces $X$ and $Y$ are homeomorphic you have to present a bijective map $f:\>X\to Y$ which is continuous in both directions. Showing that certain topological properties or invariants coincide is not enough, unless both spaces are members of a well understood family, e.g., "closed surfaces" or "open regions in ${\mathbb R}^2$".
In order to prove that two spaces $X$ and $Y$, e.g., two knots in ${\mathbb R}^3$,  are not homeomorphic you have to invoke properties or invariants that are different for the two spaces. There might be googols of bijective maps between $X$ and $Y$, and it would take an eternity to prove that each of these has a discontinuity somewhere. 
