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.
Best Answer
If $G$ is a locally finite group then every finite dimensional complex representation of $G$ is semisimple.
Suppose $U\leq V$ are finite dimensional complex representations of $G$. For each finite subset $X$ of $G$, let $S_X$ be the subset of $\operatorname{Hom}_{\mathbb{C}}(V,U)$ consisting of maps that split the inclusion map and which commute with all elements of $X$. Then $S_X$ is nonempty, by Maschke’s Theorem for the finite group generated by $X$, and is an affine subspace of $\operatorname{Hom}_{\mathbb{C}}(V,U)$.
Choose $X$ so that the dimension of $S_X$ is minimal. Then for $X\leq Y$ we must have $S_X=S_Y$. So any element of $S_X$ commutes with all elements of $G$, and so $U$ is a direct summand of $V$ as representations of $G$.