Here is Ribet's proof (expanding on Ulrich's comment):

Let $G_K:=Gal(\bar{K} / K)$ and $V_l:=T_l(E)\otimes \mathbb{Q}_l$.

The image $\rho_{l,E}(G)$ is a closed subgroup of the $l$-adic Lie group $\text{Aut}(V_l(E)) \cong \text{GL}_{2}(\mathbb{Q}_l)$ and is therefore a Lie subgroup of $\text{Aut}(V_l(E))$. Its Lie algebra $\mathfrak{g}_l$ is a subalgebra of $\mathfrak{gl}_{2}(\mathbb{Q}_l)$. We want to show that $\mathfrak{g}_l=\mathfrak{gl}(V_l)\cong \mathfrak{gl}_2(\mathbb{Q}_l)$ and the result follows.

(Note that the Lie algebra of the image $\rho_{l,E}(G_K)$ is the tangent space of the identity component of the Zariski closure of $\rho_{l,E}(G_K)$ in $\text{GL}_{2}(\mathbb{Q}_l)$. So $\mathfrak{g}_l$ `measures the representation up to finite extensions of the base field $K$', since a finite index subgroup of an algebraic group has the same identity component).

Now $V_l$ is irreducible as a $\mathfrak{g}_l$-module (this is a theorem of Shafarevich, and depends on Siegel's theorem on the finiteness of integral points on curves). Secondly, $\mathfrak{g}_l$ can't be contained in the subalgebra $\mathfrak{sl}(V_l)$ of $\mathfrak{gl}(V_l)$ since $\det(\rho_{l,E})=\chi_l$ (where $\chi_l$ is the cyclotomic character giving the action of Galois on $K^{cycl}$).

This leaves two possibilities for $\mathfrak{g}_l$: either $\mathfrak{g}_l$ is $\mathfrak{gl}_2(\mathbb{Q}_l)$ and we're done, or $\mathfrak{g}_l$ is a non-split Cartan subalgebra of $\mathfrak{gl}_2( \mathbb{Q}_l)$ (an abelian semisimple algebra coming from a quadratic field extension of $\mathbb{Q}_l$).

Faltings proved two important facts about represenations $\rho_{l,E}$:

$\rho_{l,E}$ is a semisimple representation of $G_K$ over $\mathbb{Q}_l$

$\text{End}(E)\otimes \mathbb{Q}_l \cong \text{End}_{\mathfrak{g}_l}(V_l)$.

Faltings results then rule out the possibility that $\mathfrak{g}_l$ is a non-split Cartan subalgebra of $\mathfrak{gl}_2( \mathbb{Q}_l)$ and we're done.