It is true, because $a_{n}$ strongly increase integer sequence. For any N exist k, suth that $a_{k}>N$. Then $a_{i+1}-a_{i}<\sqrt{\frac{a_{k}}{x}}$ while $i\le m, a_{m}<\frac{a_{k}}{x}$, therefore exist m, suth that $x<\frac{a_{k}}{a_{m}}<y$ if $a_{k}>\frac{xy^{2}}{(x-y)^{2}}$.

Note that, $a_{n+1}>a_n$, thus, the sequence is strictly increasing; in particular, $a_n$ diverges. Moreover, using the second part of inequality, we have, $$ a_{n+1}-a_n <\sqrt{a_n}\implies 1-\frac{a_n}{a_{n+1}}<\frac{\sqrt{a_n}}{a_{n+1}}<\frac{1}{\sqrt{a_n}}\implies \frac{a_n}{a_{n+1}}>1-\frac{1}{\sqrt{a_n}}. $$In particular, since $a_n$ diverges, for any given $\epsilon>0$, there exists an $N$ such that, for every $n\geq N$, $$ \frac{a_n}{a_{n+1}}>1-\epsilon. $$We will now argue through contradiction. Suppose that, there exists a pair $0<x<y<1$, for which, for every $k,m$; $$ \text{either}\quad \frac{a_k}{a_m}<x \quad\text{or}\quad \frac{a_k}{a_m}>y. $$Now, we select an $\epsilon>0$, for which, $$ \epsilon<1-\frac{x}{y}<1-x. $$The reasoning will become clear soon. First, observe that this indeed makes sense, since, $x<x/y$ as $y<1$. Now, for this epsilon; select $N_\epsilon$ such that, $$ n\geq N_\epsilon \implies \frac{a_n}{a_{n+1}}>1-\epsilon>x. $$Since, $1-\epsilon>x$; we have $\frac{a_n}{a_{n+1}}>x$; hence, $$ \frac{a_n}{a_{n+1}}>y. $$We now keep multiplying, $$ \frac{a_n}{a_{n+1}}\cdot \frac{a_{n+1}}{a_{n+2}}\cdots \frac{a_{n+t-1}}{a_{n+t}}=\frac{a_n}{a_{n+t}}<\frac{a_n}{a_{n+1}}. $$In fact, as $t\to \infty$, $\frac{a_n}{a_{n+t}}\searrow 0$. Hence, one should eventually go below $y$; and therefore, by hypothesis, also below $x$. Let, $k$ be the first time when this holds, namely, $$ \frac{a_n}{a_{n+k-1}}>y \quad \text{but}\quad \frac{a_n}{a_{n+k}}<x. $$Observe that, $$ x>\frac{a_n}{a_{n+k}}=\underbrace{\frac{a_n}{a_{n+k-1}}}_{>y}\underbrace{\frac{a_{n+k-1}}{a_{n+k}}}_{>1-\epsilon}>y(1-\epsilon) \implies \epsilon>1-\frac{x}{y}, $$contradicting with the selection of $\epsilon$.