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Well, the simple groups ${\rm SL}(2,2^{n})$ ( $n >1$) certainly have $2$-dimensional irreducible representations in characteristic $2$, and in fact $A_{5} \cong {\rm SL}(2,4)$ (to see this, just note that ${\rm SL}(2,4)$ has order $(4+1)(4^{2}-4) = 60 ).$ The standard argument to show that a non-Abelian finite simple group can't have a $2$-dimensional irreducible representation does not work in characteristic $2$, since it relies on the fact that an involution (which must be non-central using simplicity) must have determinant $-1$, since one of its eigenvalues must be $1$ and the other $-1.$ But again since $G$ is non-Abelian simple, every representation has trivial determinant, a contradiction. But this line argument does not work over a field of characteristic $2$.