Upon square roots' difference I have to prove this statement: $$\forall a\in\mathbb R\ \exists m,n\in\mathbb N^*:a<\sqrt{m}-\sqrt{n}<a+0.01$$
For example, given $a=0$, $m=50^2+2$ and $n=50^2+1$ satisfy these inequalities since $$m>n\rightarrow \sqrt m-\sqrt n>0$$ $$\sqrt m-\sqrt n=\frac 1{\sqrt m+\sqrt n}<\frac 1{2\sqrt n}<\frac1{100}$$
but the proof for the general situation eludes me... Ideas?

Let's have it in a more convinient form
$$a<\sqrt{m}-\sqrt{n}<a+\varepsilon.\tag{1}$$
Then we extend the inequality to $$a-\varepsilon<\sqrt{m}-\sqrt{n}<a+\varepsilon.\tag{2}$$
Indeed, we can always obtain (1)-form from (2)-form with $(a'-\varepsilon',a'+\varepsilon')\to (a,a+\varepsilon)$ (it's a linear transform $a'=\frac\varepsilon 2+a, \,\varepsilon'=\frac\varepsilon 2$).
$$a-\varepsilon+\sqrt{n}<\sqrt{m}<a+\varepsilon+\sqrt{n}\tag{3}$$
$$(a-\varepsilon)^2+2(a-\varepsilon)\sqrt{n}+n<m<(a+\varepsilon)^2+2(a+\varepsilon)\sqrt{n}+n\tag{4}$$
Now we want $$(a+\varepsilon)^2+2(a+\varepsilon)\sqrt{n}+n-
\left((a-\varepsilon)^2+2(a-\varepsilon)\sqrt{n}+n\right)>1\tag{6}$$ so there at least one integer $m$ within the interval
$$\left(
(a-\varepsilon)^2+2(a-\varepsilon)\sqrt{n}+n,
(a+\varepsilon)^2+2(a+\varepsilon)\sqrt{n}+n
\right)$$
or even ti's better to have at least $2$ integers for the case when one of them is equal to $n$. I'll show for $1$, for $2$ it's quite similar. (6) is equivalent to
$$4a\varepsilon+4\varepsilon\sqrt{n}>1\tag{7}$$
$$a+\sqrt{n}>\frac{1}{4\varepsilon}\tag{8}$$
So it's sufficient to take $$n>\left(\frac{1}{4\varepsilon}-a\right)^2\tag{9}$$
Now, we can obtain, for example, an approximation for $\pi$ with $10^{-9}$ precision:
$$\sqrt{62500000000000001}-\sqrt{62499998429203684}\approx \color{blue}{3.141592653}7392087789$$
