Combinatorial proof that$|ma-\sqrt{p_n}|$ get's arbitrarily small for infinitely many $n$ Given a sequence of positive numbers, $\varepsilon_n$, which converge to zero, there is some a>0 such that for infinitely many n there is an m for which
$$|ma-\sqrt{p_n}|<\varepsilon_n$$
where $p_n$ is the nth prime number. I have a proof of this fact produced below, but it comes as an edge case of a lemma based on Baire's theorem. Is there a more direct approach which uses number theory or combinatorics?
Lemma: Let $f\colon \mathbb R \to \mathbb R$ be a continuous function such that for all $a>0$ we have $\lim\limits_{n\to \infty} f(na)=0$. Then $\lim\limits_{x\to \infty} f(x)=0$.
We prove this as follows: Choose any $\varepsilon>0$ and set $$E_n = \{x\colon |f(mx)|\leq \varepsilon \text{ for all }m\geq n\}=\bigcap\limits_{m\in \mathbb N} \{x\colon mx\in f^{-1}([-\varepsilon,\varepsilon])\}.$$ Each of these sets will be closed by the continuity of $f$, and by the definition of our function we have that $\mathbb R_+ = \bigcup\limits_{n\in \mathbb N} E_n$, so that by Baire's theorem at least $E_n$ has nonempty interior.
Thus we have an interval $I=[a,b]$, and some positive integer $n_0$ such that for all $n>n_0$, $f(nI)\subset [-\varepsilon,\varepsilon]$. We also have that if $n$ is big enough that $na<(n-1)b$ so that integer multiples of this interval will eventually cover the tail of the $\mathbb R_+$. All $x$ large enough will then be in a large integer multiple of $I$, and so we have $f(x)\in[-\varepsilon,\varepsilon]$. QED
Now to prove the statement in the question, take $f$ to be $1$ at the square root of prime numbers, $0$ on $\mathbb R_+ \setminus \bigcup (\sqrt{p_n}-\varepsilon_n,\sqrt{p_n}+\varepsilon_n)$ and linear elsewhere. This function does not go to $0$ as $x\to \infty$ so we must have the existence of at least one point $a$ such that
$$\lim_{m\to \infty}f(ma)\not\to 0.$$
It follows that $ma$ must wind up in infinitely many of the peaks centered about $\sqrt{p_n}$.
 A: This can be done using a standard "satisfying the $k$th condition at the $k$th step" argument.
Let's prove that there exists a sequence of, for each natural number $k$, natural numbers $n_k$ and $m_k$ and a closed interval $I_k$ of nonzero length, such that $I_{k+1}\subseteq I_k$ for all $k$, and such that for each $k$, $m_k I_k \subseteq (\sqrt{p_n} -\epsilon_n, \sqrt{p_n}+\epsilon_n)$.  Then taking $a$ to be any number contained in $I_k$ does the job. 
To prove this, we can inductively assume we have this data up to $k$, and try to construct $n_{k+1}, m_{k+1}, I_{k+1}$.
Lemma: For each interval $I_k$ in the positive reals of nonzero length, there are infinitely many $n,m$ with $\sqrt{p_n} \in m I_k$. 
Proof: The primes are unbounded, and as you note, all sufficiently large numbers are contained in $m I_k$ for some $m$, so all sufficiently large prime square roots are contained in $m I_k$ for some $m$.
So we can find an $m$ and an $n$, not already picked, with $\sqrt{p_n} \in m I_k$. Then the intersection of $m I_k$ with $(\sqrt{p_n} -\epsilon_n, \sqrt{p_n}+\epsilon_n)$ contains a closed interval of nonzero length. Take $n_{k+1} = n$, $m_{k+1} = m$, and $I_{k+1}$ to be this interval divided by $m$, which is contained in $I_k$ and also satisfies $m_{k+1} I_{k+1} \subseteq (\sqrt{p_n} -\epsilon_n, \sqrt{p_n}+\epsilon_n)$.
