I have finished writing the proof and I'm afraid I made a mistake on the way because I'm not sure I can deduce the bottom line of the proof.
Here it goes:
Let us examine the action of $G \mapsto G$ on itself by multiplication from the left.
This action induces a homomorphism $\phi: G \to S_{|G|}$. From Cayley's theorem we have $G \cong Im\phi \le S_{4n+2}$.
Let us mark $H = Im\phi$. We know $A_{4n+2} \lhd S_{4n+2}$ and from Diamond theorem we have $H \cap A_{4n+2} \lhd H$.
Let us mark $H \cap A_{4n+2} = N$. We want to show $N \lt H$ and also $N$ isn't trivial.
From Cauchy's theorem for $p=2$: exists $g\in H$ such that $|g| = 2$.
Let us assume $g$ is constructed by foreign cycles marked $c_i$ for $1 \le i \le m$ for some $m \in \mathbb N$.
Then we have $g = c_1 ... c_m$. We know $|g| = {\rm lcm}(c_i)$ and since $|g| = 2$ we have $\phi(g)$ is a product of 2-cycles.
Notice $|g|=2 \implies g \neq e$ and therefore for all $g^{'} \in G$ we have $gg^{'} \neq g$ meaning $\phi(g)$ doesn't "leave" any object in it's place.
Therefore we have $\phi(g)$ is a product of exactly $2n+1$ 2-cycles and this implies $\phi(g) \notin A_{4n+2}$ and therefore $N \lt A_{4n+2}$.
Now notice, for all $n \in \mathbb N$ exists some $p>2$ primal (and obviously odd) such that we have some $e \neq g \in H$ such that $|g| = p$.
In this case we have $\phi(g)$ is a product of oddsized-cycles and therefore $\phi(g) \in A_{4n+2}$ and therefore $N$ isn't trivial.
This implies $N \lhd H$ a proper subgroup.
Now this is the part I'm not sure I'm allowed to deduce
Since we know $G \cong H$ we have some $H^{'} \lt G$ such that $H^{'} \cong N$ therefore meaning $H^{'} \lhd G$ a proper subgroup, thus implying $G$ isn't simple.
Another problem
I feel like I'm proving a stronger theorem since I could've used all the statements about $4n+2$ for every even number? or maybe I'm just mixed up. Thanks in Advance!
To react to your remark "Is the current proof all right now"? First of all, the Cayley embedding induces a homomorphism $\phi: G \to S_{|G|}$ (and not to $S_{|G\backslash H|}$). This is an injection so the First Isomorphism Theorem guarantees that $G \cong \phi(G)=H$. Now, as you noted $H \cap A_{4n+2} \unlhd H$. Assume that we have equality here, then $H \cap A_{4n+2} = H$, which is equivalent to $H \subseteq A_{4n+2}$.
You then observe rightly that by Cauchy $G$ has an element of order $2$, hence $\phi(g)$ is an element of order $2$. Then you describe $\phi(g)$ correctly: it is a product of $2n+1$ transpositions ($2$-cycles), hence an odd permutation, so $\phi(g) \notin A_{4n+2}$, a contradiction. We conclude that $H \cap A_{4n+2}$ is a proper normal subgroup of $H$. Your argument for being non-trivial is correct. The part you are not sure about is also correct. The point is that you again can use the First Isomorphism Theorem and look at $\phi^{-1}(H \cap A_{4n+2})$, being a non-trivial normal subgroup of $G$ (remember that the map $\phi: G \rightarrow \phi(G)$ is an isomorphism!)
So what did you learn: the whole motivation behind the proof is embedding/representing $G$ in another group where you have more tools or control to study $G$. This happens more oftenly in mathematics: represent your structure in something bigger where more tools are available: think of calculating difficult real integrals by interpreting them over $\mathbb{C}$. Later you might hit representation theory of finite groups where you represent them for example in the linear groups $GL(n,\mathbb{C})$ and suddenly you can apply linear algebra, Galois theory and (algebraic) number theory to prove nice things. Burnside used this last "trick" to prove that any group of order $p^aq^b$ ($p$, $q$ primes) is solvable. No need for the sledge hammer of Feit-Thompson for $p, q$ odd (which did not yet exist at that time!)