Without using Sylow: Group of order 28 has a normal subgroup of order 7

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.

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 $\lvert N\rvert = 7$.