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.
Since $|G|=10$, therefore for $g \in G$ the possible orders are $1,2,5,10$. Suppose
- There is an element $a \in G$ such that $|a|=10$, then $|a^2|=5$, thus we have an element of order $5$.
- Suppose there is an element of order $5$, then we have nothing to prove.
- So what if all the elements have order $2$? Then you can show that the group is abelian (small exercise). Now consider distinct elements $a,b \in G$ such that $|a|=|b|=2$, then $|ab|=2$, hence $H=\{e,a,b,ab\}$ will form a subgroup of $G$. But $|H|=4$ and $4 \not| 10$. It violates Lagrange's theorem. So such a $H$ cannot exist. That means not all the elements can have order $2$.
I hope this resolves it within the scope of things you know at this stage.
Best Answer
By Cauchy's theorem the group, say $G$, has an element $g$ of order $7$. To the contrary, assume $G$ has an element $h$ of order $7$ not contained in $\langle g \rangle$. Then, $\langle g \rangle \langle h\rangle$ has size $49 > 28$ (why?), a contradiction.