any $2$-dimensional rep of a finite, non-abelian simple group is trivial

Okay, we're going to have to use some heavy artillery to start off, but I can't think of another way to begin.

Suppose $\rho: G\to \text{GL}_{2} (\mathbb{C})$ is nontrivial. Observe that since $G$ is simple and the representation is nontrivial, we must have $\text{ker} \, \rho =\text{ker}\, \chi = (e)$ (where $\chi$ is the character of this representation). The Feit-Thompson Theorem (!!!) tells us $|G|$ is even. By Cauchy's Theorem, $G$ must have an element $x$ of order $2$.

Now, define $$\hat{\rho}: G \to \text{GL}_{1} (\mathbb{C}) \cong \mathbb{C}^{\times}$$ by $\hat{\rho}(g) = \text{det} (\rho(g))$. Evidently, $\hat{\rho}$ is a homomorphism, hence it gives a degree 1 representation of $G$. We know this representation must be trivial. In other words, $\text{det} (\rho(g)) = 1$ for all $g\in G$. That said, we also know that $\rho(x)^2 = \text{Id}$. The set of eigenvalues of $\rho(x)$ is either $\{1, 1\}$, $\{1,-1\}$, or $\{-1,-1\}$. The first possibility is out of the question, since $\text{ker} \chi = (e)$. The second possibility cannot occur, since then $\text{det} (\rho(x)) = -1$. Thus, the eigenvalues of $\rho(x)$ are $\{-1, -1\}$. The characteristic polynomial of $\rho(x)$ is $(X+1)^2$, and $\rho(x)$ also satisfies $X^2 - 1$. Since the minimal polynomial of $\rho(x)$ must divide both of these, it follows $\rho(x)$ satisfies $X+1$, i.e. $\rho(x) = -\text{Id}$.

Lastly, since $\rho(x)$ is a scalar multiple of the identity, it commutes with any matrix. In particular, for any $g\in G$, we have $$\rho(g) \rho(x) = \rho(x) \rho(g) \implies \rho(gxg^{-1} x^{-1}) = \text{Id}$$ Triviality of $\text{ker} \, \rho$ implies $gxg^{-1} x^{-1} = e$ for all $g\in G$, hence $x\in Z(G)$. Accordingly, $Z(G)$ is a nontrivial normal subgroup of $G$, so it must equal $G$. But $G$ is non-abelian by assumption.

Using a lot less machinery:

Such a representation must land in $SL_2(\mathbb C)$, and finite subgroups of $SL_2(\mathbb C)$ are completely classified (they are either cyclic, dihedral, tetrahedral, octahedral or isocahedral). None of these groups are simple, unless cyclic of prime order.

One can prove this classification by letting a finite subgroup $G$ of $SL_2(\mathbb C)$ act on the Riemann sphere $\mathbb P^1$. One can compare the Riemann-Hurwitz formula for the morphism $\mathbb P^1 \to \mathbb P^1/G$ to the class equation of $G$ to deduce bounds on the individual terms appearing in the class equation, thus narrowing down the possible groups.

This is exercise 3.3 in Character Theory of Finite Groups by Isaacs. It is easily deduced from the following facts:

• Because $G$ is perfect, $\chi$ has image in $SL(2,\mathbb{C})$;
• Because $\chi$ is irreducible, $\chi(1)$ divides $|G|$;
• The only element of order $2$ in $SL(2,\mathbb{C})$ is $-I$ (which is central).