In general, a (smooth, complex, finite-dimensional) representation of a compact Lie group (such as SO(n)) decomposes into a direct sum of irreducible representations. The irreducible representations of SO(n) have been classified; this is the classical theory of spherical harmonics.
For SO(3), it's easy to work things out by hand. Recall that we have the 2:1 covering SU(2) -> SO(3). So, given an irred rep SO(3) -> GL(V), we can compose it with the covering map to get a rep SU(2). This SU(2) rep is easily seen to be irreducible. The important thing to notice is that the kernel {±I} of the covering map will act trivially on V. Conversely, whenever you have an irreducible representation of SU(2) in which -I is acting trivially, you get an irreducible representation of SO(3).
So now we have to determine the irred reps of SU(2), but this is pretty straightforward (and can be found in virtually every textbook on the rep theory of Lie groups). One approach would go as follows. Because SU(2) is simply connected, its representations lie in one-to-one correspondence with representations of its Lie algebra su(2), which in turn Lie in one-to-one correspondence with representations of the complexification ##\mathfrak{sl}_2\mathbb C = \mathfrak{su}(2) \otimes \mathbb C##. The irred reps of sl_2(C) are ##V_n = \text{Sym}^n \mathbb C^2## (##n=0, 1, 2, \ldots##), where sl_2(C) acts on C^2 in the obvious way. [Note that dim V_n = n+1.] (See, e.g., the chapter in Fulton & Harris on sl_2.)
From this description we can see that ##-I \in SU(2)## acts on V_n by ##(-1)^{n}##, i.e., -I acts trivially on V_n iff n is even (in which case dim V_n is odd).
Consequence: SO(3) has no irreducible representations representations of even dimension.
It follows that a 2-dimensional representation of SO(3) will decompose into a direct sum of one-dimensional representations, which are necessarily trivial (by, e.g., the above description).
