Without using Sylow: Group of order 28 has a normal subgroup of order 7 
Prove that a group of order 28 has a normal subgroup of order 7.

How can I prove this without using Sylow's theorem?
I know by Cauchy’s theorem, there exists an $x\in G$ with order 7, now I just need to prove it has a normal subgroup. 
 A: We have $\lvert G\rvert = 28 = 2^2\cdot 7$.
Let $a_7$ be the number of $7$-Sylow groups in $G$.
By Sylow
$$a_7\equiv 1\mod 7\qquad\text{and}\qquad a_7 \mid 4\text{.}$$
This implies $a_7 = 1$, so there is a unique subgroup $H$ of $G$ of order $7$.
For all $g\in G$, $gHg^{-1}$ is again a subgroup of order $7$, which forces $gHg^{-1} = H$ for all $g\in G$. So $H$ is a normal subgroup of order $7$.
Edit
Only now I realized that you don't want to use the Sylow theorems. Here is an alternative, following the hint of @Tobias Kildetoft:
Since $7$ is prime, by Cauchy $G$ contains an element of order $7$.
It generates a subgroup $H$ of $G$ of order $7$.
Look at the group action of $G$ on the set $G/H$ of left-cosets of $H$ by left multiplication. Obviously, this group action is transitive. Because of $\lvert G/H\rvert = 28/7 = 4$, it gives rise to a group homomorphism
$$
\varphi : G \to S_4
$$
Now look at $N = \ker(\varphi)$, which is a normal subgroup of $G$.
From the transitivity of $\varphi$, we get $\operatorname{im}(\varphi) \geq 4$, which translates to $\lvert N\rvert \leq 28/4 = 7$.
By the homomorphism theorem, $G/N \cong \operatorname{im}(\varphi)$, so $\lvert G\rvert/\lvert N\rvert$ divides $\lvert S_4\rvert = 24$. This implies $28 \mid 24 \lvert N\rvert$, so $7\mid\lvert N\rvert$.
So the only remaining possitiblity is $\lvert N\rvert = 7$.
A: Another different proof: since as you said you know that $G$ has an element of order $7$, by Cauchy, you also know that there's at least one subgroup of order $7$, let's call it $H$.
Suppose that $K \leq G$ is another subgroup of order $7$, then we can consider the subset $HK$ that has order $|HK|=|H||K|/|H \cap K|$.
If $H$ and $K$ were distinct then $H \cap K$ should be a proper subgroup of both of them, but since they have order the prime $7$ this is possible iff $H \cap K=(\text{id})$ and so $|HK| =7\cdot 7 = 49$ which is clearly bigger then $28$.
We arrived to an absurdity, so we have to conclude that $H$ is the only subgroup of order $7$ and so it's characteristic, hence normal.
Edit (more details): let's consider a generic automorphism $\varphi \colon G \to G$, then by properties of homomorphisms $\varphi(H)$ is a subgroup of $G$ and since $\varphi$ is bijective $\varphi(H)$ should have order $7$.
Because as we have proved $H$ is the only subgroup of order $7$ it follows that $\varphi(H)=H$: so $H$ is fixed by all the automorphisms, i.e. is characteristic.
From this follows normality since a subgroup is normal iff is fixed by all inner automorphisms, i.e. is fixed by all the automorphisms of the form
$$x \mapsto gxg^{-1}$$
for some $g \in G$.
Since $H$ is fixed by every automorphism it's fixed in particular by the inner automorphism and so it's normal.
A: Another kind of proof. First, we remark that if $G$ is abelian, we can apply Cauchy to get a group of order $7$, which must be normal.
Consider the conjugacy classes of a non-abelian $G$. They can only have sizes $1,2,4,7,14,28$. We know the identity is in a class by itself, so we have $27$ more elements to divvy up. We would like to show one of them has conjugacy class $1$, $2$, or $4$.
Clearly, we cannot form $27$ out of just $7$s and $14$s, so at least one of those three is required, call it $k$. $G$ acts on the elements of that conjugacy class by conjugation, so it induces a homomorphism $\varphi$ from $G$ into $S_k$. But $|S_k|$ is at most $24$, which is less than $28$, so $\ker \varphi$ is non-trivial. Since $G$ is non-abelian, $\ker \varphi \ne G$, so we have our normal subgroup, $\ker \varphi$.
This works for groups of order $mn$, where $m$ and $n$ are relatively prime and $n \nmid (m -1)!$.
