Let me start by saying that I'm a student self-studying this for fun so I might be asking a really silly question here, anyways:
According to Wikipedia, a topological invariant is defined as a property of a topological space that remains invariant under homeomorphisms. Obviously, this definition is correct, but I'm failing to see how this definition is not redundant since a homeomorphism already preserves the features of a topological space. Can someone please clarify?
Your statement a homeomorphism already preserves the features of a topological space is rather vague --- it does not tell us what features matter and what features don't matter. The definition of a topological invariant is not vague, it gives us a way to say exactly what kinds of features matter.
Suppose we have a particular feature, a property, of certain topological spaces. We want to know: Is this property a topological invariant? The great advantage of the definition of a topological invariant is that you can apply it to translate the question into a precise, answerable, mathematical question (maybe the question is difficult to answer, but it is at least precise enough that you can attempt to answer it by mathematical methods).
Let's take an example. Given a subset $X \subset \mathbb R^n$, let us say that $X$ is bounded if there exists a real number $B>0$ such that for all $x,y \in X$ we have $|x-y| \le B$. We want to know:
Is boundedness of subsets of $\mathbb R^n$ a topological invariant?
Translation:
If two subsets of $\mathbb R^n$ are homeomorphic, is it true that one is bounded if and only if the other is bounded?
Answer:
No, because there is a counterexample. In the case $n=1$, we have a bounded subset $(-1,+1)$ and an unbounded subset $\mathbb R$ and these two subsets are homeomorphic to each other: the function $f : (-1,+1) \to \mathbb R$ defined by $f(t) = \arctan(t \, \pi \, / \, 2)$ is a homeomorphism.
Let's take another example. Let us say that $X \subset \mathbb R^n$ is closed if for every sequence $(x_n)$ in $\mathbb R^n$, if $(x_n)$ converges to a limit $p \in \mathbb R_n$, and if $x_n \in X$ for all $n$, then $p \in X$. We want to know:
Is closedness of subsets of $\mathbb R^n$ a topological invariant?
Translation:
If two subsets of $\mathbb R^n$ are homeomorphic, is it true that one is closed if and only if the other is closed?
Answer:
No, and the same two subsets as before give a counterexamples: the subset $(-1,+1) \subset \mathbb R$ is not closed, whereas the subset $\mathbb R \subset \mathbb R$ is closed.
And here's the punch line example. Let us say that $X \subset \mathbb R^n$ is compact if it is closed and bounded. We want to know:
Is compactness of subsets of $\mathbb R^n$ a topological invariant?
Translation:
If two subsets of $\mathbb R^n$ are homeomorphic, is it true that one is compact if and only if the other is compact?
It seems hopeless; "closedness" is not a topological invariant; "boundedness" is not a topological invariant; how could "closedness and boundedness" be a topological invariant? Interestingly, however, the previous counterexample does not work: $(0,1)$ is bounded but not closed; whereas $\mathbb R$ is closed but not bounded. And as one learns in elementary topology, compactness is indeed a topological invariant; this is usually proved in the first weeks of an elementary topology course.
You say you are heavily interested in Chern classes. There are other kinds of classes of interest in algebraic topology: Pontryagin classes; Euler classes. And associated to those are certain numbers: Chern numbers; Pontryagin numbers; the Euler number also called the Euler characteristic.
Let me simplify matters by focussing on the Euler characteristic for my final example, which aligns more closely with the concept of topological invariance.
We want to know:
Is the Euler characteristic a topological invariant?
Let's translate this carefully. Define the Euler characteristic $\chi(X)$ of a finite simplicial complex $X$ to be the alternating sum of the numbers $k_n$ of simplices in each dimension $n$: $$\chi(X) = \sum_n (-1)^n k_n $$
If $X$ and $Y$ are simplicial complexes, and if $X$ is homeomorphic to $Y$, is the Euler characteristic of $X$ equal to the Euler characteristic of $Y$?
Answer: Yes it is, but the proof is hard. It requires around a semester of algebraic topology to understand: one needs to understand topological invariance of simplicial homology; and then one needs to prove a formula for the Euler characteristic expressed as the alternating sum of the ranks of the simplicial homology groups.
So if you ever get around to thinking about a really difficult proof of topological invariance, namely the invariance of Pontryagin numbers (which are indirectly related to Chern classes), perhaps you'll appreciate better how a precise definition of topological invariance leads to a precise and answerable statement.
You’re running into a problem of circular definitions because it appears, in your head, that the definition of homeomorphism is something which preserves features of topological space. Homeomorphism is a technical term defined by talking about pre-images of open sets. Once you define that, then you can define a topological invariant as something that remains invariant under any homeomorphism.
