Here's one way to prove that there is just one subgroup without thinking too much about it. You have an abelian group $G$ (the particular primes involved in the factorisation of $|G|$ don't matter) and you want to find how many $p$-subgroups of some type it contains. Since $G$ has a unique Sylow $p$-subgroup $P$, every such $p$-subgroup is contained in $P$ thus the problem translates into one about counting $p$-subgroups of given type inside an abelian $p$-group.
I wrote all that just to say that it suffices to count the number of subgroups of $P = C_{25} \times C_{125}$ of type $C_{25} \times C_{25}$. Ignoring the particular prime $p=5$, we want to count the number of subgroups of $P = C_{p^2} \times C_{p^3}$ of type $C_{p^2} \times C_{p^2}$.
There is a standard theorem which allows you to do that. Call $H$ of type $\mu = (\mu_1,\ldots,\mu_k)$ if $H= C_{p^{\mu_1}} \times \ldots \times C_{p^{\mu_k}}$. The theorem says that the number of subgroups of type $\nu = (\nu_1,\ldots,\nu_\ell)$
in an abelian $p$-group of type $\lambda = (\lambda_1, \ldots, \lambda_\ell)$ is given by the formula
$$\prod_{i\geq 1} p^{\nu_{i+1}'(\lambda_i' - \nu_i')} {\lambda_i' - \nu_{i+1}' \brack \nu_i' - \nu_{i+1}'}_p,
$$
where $\lambda', \nu'$ are the conjugates of the partitions $\lambda$ and $\nu$, respectively, and
\begin{equation}
{n \brack k}_p = \prod_{i=0}^{k-1} \frac{1 - p^{n-i}}{1-p^{i+1}}
\end{equation}
is the number of $k$-dimensional subspaces of an $n$-dimensional vector space over the field $\mathbb{Z}/p\mathbb{Z}$;
see, for instance, this (Eqn. ($1$)).
In our example, we have $\nu =(2,3)$, $\nu' = (2,2,1)$ and $\lambda=\lambda'=(2,2)$. If you do the algebra you'll see that the answer is $1$.
This whole thing is very likely overkill, but at least it allows you to solve this and other similar problems uniformly.
