Let $G$ be a group of order $p^2$ and put $\mathcal A=\{U\leq G, \#U=p\}$.
What is $\#\mathcal A$?
If $G$ is cyclic, then $G$ is generated by some element $x$ of order $p^2$. It seems like there is only one order $p$ subgroup here.
Question 1 How do I prove this rigorously?
If $G$ is not cyclic, then there is no element of order $p^2$. By Lagrange's theorem, any nontrivial element $g\in G$ then has order $p$, hence generates an order $p$ subgroup $\langle g \rangle\in \mathcal A$.
Question 2 Which of these groups $\langle g\rangle$ coincide?
Question 3 Are there any more order $p$ subgroups?
Since we know
$$G\cong C_{p^2}\;\;\;\text{or}\;\;\;G\cong C_p\times C_p$$
there are not many options: in the first case $\; |A|=1\;$ as any cyclic group of finite order has one single subgroup of any order dividing the group's.
In the second case: since we can consider $\;G\;$ a a vector space of dimension two over the prime field of characteristic $\;p\,,\,\,\Bbb F_p\;$, the wanted number equals the number of different subspaces of dimension one this space has. Can you take it from here?