Proof

We prodeed by induction on the order of $A$. For ${\vert A \vert} = 2$ we have that $A = \mathbb{Z}_2$ is the unique group of order 2 and the statement holds for $p =2$.

Assume then that the statement has been show for groups of order $\lt n$ and let ${\vert A \vert} = n$.

If $A$ has no non-trivial proper subgroup then $n$ must be prime and $A = \mathbb{Z}_n$ a cyclic group and the statement follows.

If $A$ does have a non-trivial proper subgroup $H \hookrightarrow A$ then $p$ divides either ${\vert H \vert}$ or $\vert A/H\vert$.

In the first case by induction assumption $H$ has an element of order $p$ which is therefore also an element of $G$ of order $p$.

In the second case there is by induction assumption an element $a \in A$ such that $a + H \in A/H$ has order $p$. Since the order of $a + H \in A/H$ divides the order of $a \in A$ it follows that $a$ has order $k p$ for some $k \in \mathbb{N}$. Then $k a$ has order $p$.