I'm trying to learn some Lie algebra without much knowledge of representation theory. While being asked to prove some things about an sl(2)module, why can one assume that the module is irreducible and has dimension 2?

The representation theory of sl(2) is the basis for understanding the representation theory of any Lie algebra, and it is not difficult at all. An extremely readable account of it is given in chapter 11 of Fulton and Harris' "Representation Theory, a First Course". 


You can't assume the module is irreducible, since many aren't! However, if you want to learn something about modules of $sl_2$ it helps to make the following observations:
In particular, when you'll be solving 2, you'll notice that there is exactly one irreducible modle of dimension 2. In fact, there is one of dimension 3, one of dimension 4, etc. 


The irreducible modules of sl2 are of the form Sym^r(\C^2), i.e homogeneus polynomials of degree r and 2 variables x,y. So you have irreducible modules of dimension 1,2,3,4,.... etc. Even more, sl2 acts on these spaces by x\diff y, y\diff x and the operator (x\diff y)(y\diff x)(y\diff x)(x\diff y). From this it's easy to calculate the vectors with maximal weights in each irreducible representation. 

