Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to $R^n$? Timothy Gowers asks Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to $R^n$? and lists some reasons. The most powerful of these is probably 

There are many important examples throughout mathematics of infinite-dimensional vector spaces. If one has understood finite-dimensional spaces in a coordinate-free way, then the relevant part of the theory carries over easily. If one has not, then it doesn't.

I mean sure, but what else? Does anyone know examples of specific vector spaces?

For any integer $k$, the set $M_k$ of complex-differentiable functions $f$ defined on the upper-half plane $\{x+iy: \, y > 0\}$ that satisfy the equations $$f(z+1) = f(z), \; \; f(-1/z) = z^k f(z)$$ and have limit $\lim_{y \rightarrow \infty} f(iy) = 0$ is a vector space over $\mathbb{C}$. 
Two specific elements of $M_k$ include the functions $$E_4(z) = 1 + 240 \sum_{n=1}^{\infty} \sigma_3(n) e^{2\pi i n z} \in M_4$$ and $$E_8(z) = 1 + 480 \sum_{n=1}^{\infty} \sigma_7(n) e^{2\pi i nz} \in M_8.$$ Here, $\sigma_k(n)$ is the divisor sum $\sum_{d | n} d^k$.
Assuming that $E_4 \in M_4$ it is rather easy to show that $E_4^2 \in M_8.$
It can be proved that $M_8$ is one-dimensional, so $E_4^2$ is a multiple of $E_8$. Comparing constant coefficients tells you that they must be equal, and comparing the others gives you the formula $\sigma_7(n) = \sigma_3(n) + 120 \sum_{m=1}^{n-1} \sigma_3(m) \sigma_3(n-m).$
For example $$\sigma_7(2) = 1 + 2^7 = 1 + 2^3 + 120$$ and $$\sigma_7(3) = 1 + 3^7 = 1 + 3^3 + 120(1+2^3 + 1 + 2^3).$$
A lot of vector spaces like this show up in number theory. They are typically finite-dimensional but working out a basis is pretty hard (certainly harder than showing that they are finite-dimensional).
