I think it is the characteristic. $A_5$ naturally sits in $\text{SO}(3)$ as the group of rotational symmetries of the icosahedron. Pulling $A_5$ back along the double cover $\text{SU}(2) \to \text{SO}(3)$ gives a double cover of $A_5$, the <a href="http://en.wikipedia.org/wiki/Binary_icosahedral_group">binary icosahedral group</a> $G$, and this group is therefore equipped with a $2$-dimensional complex representation. 

By <a href="http://mathoverflow.net/questions/847/is-any-representation-of-a-finite-group-defined-over-the-algebraic-integers">this MO question</a>, this representation is defined over $\mathcal{O}_K$ for a number field $K$, so we can reduce it $\bmod 2$; the double cover $G \to A_5$ has kernel $\{ \pm I \}$ which is killed when working $\bmod 2$, so we get a $2$-dimensional representation of $A_5$ in characteristic $2$. 

I am not sure exactly what proof you're referring to of the fact that a simple group $G$ can't have an irreducible representation of dimension $2$. I hope it is this one:

- The dimension of an irreducible representation divides the order of $G$.
- By Cauchy's theorem, $G$ has an element of order $2$.
- By averaging, $G$ is a subgroup of $\text{U}(2)$. By simplicity, it is a subgroup of $\text{SU}(2)$.
- But the only element of order $2$ in $\text{SU}(2)$ is $-I$, which is central; contradiction.

I don't know if the first step works in characteristic $2$, and the third step is problematic also (we should work with $\text{SL}_2$), but the last step definitely fails (as seen above).