There must be some $a\in G$ such that 7 does not divide $ |G| \over |C_G(a)| $
For the first thing, note that LHS is divisible by $7$. If all the terms in the summation were divisble by $7$ then, the RHS, will be $1 \mod 7$. So, there must be atleast one $x$ with the property you claim.
It follows that ${|G| \over |C_G(a)|} = 2 $ or $4$
Note that, $I=\dfrac{|G|}{|C_G(x)|}$, called the index of $C_G(x)$ in $G$, is a number that divides the order of the group. (Why?)
So, if $I$ is not $7$, it must be $2$ or $4$. (It cannot be $1$ by the very definition of class equation.)
$\Rightarrow [G:C_G(a)] = 2$ or $4$ $\Rightarrow 28 |2!$ or $4!$.
The conclusion in the last line is a very important lemma due to Poincare. Let $G$, a simple group act on $G/H$, where $H$ is a proper subgroup of $G$. This gives you a homomorphism from $G \to \operatorname{Sym}(G/H)$. Is the action trivial? Now what will its Kernel be? What does the first isomorphism theorem tell you?
P.S.: I quite disagree with the sequel, especially in the last part. You start with the assumption that group is simple. Then you arrive at a contradiction. *"Therefore group of order 28 is simple." * is placed wrongly in your proof.
A nicer way would be to:
Suppose $G$ is a simple group. Then, its centre is trivial.
Because, if this was non-trivial, this will be a non-trivial normal subgroup of $G$. If it is proper, we are done because this is a contradiction to the "Simpleness" of the group. If it is improper, then, the group is abelian. By Cauchy's Theorem for Abelian groups, there must be an element of order $2$ which should be normal (because any subgroup in $G$ will be normal).
Now carry out the argument I have showed you almost explicitly!