Conjecture regarding trapping rational numbers in some special intervals Conjecture:
Let $b\in\mathbb{N}_{\geq3}$ and $\{x_i\}$ be a collection of $b−2$ rational numbers greater than $1$. Does there always exist a natural number $a$ such that for all $i$ there exists some natural number $1\leq c\leq a$ such that: 
$$\displaystyle \frac{ab-1}{(c-1)b+1}\geq x_i \geq \frac{ab-1}{cb-1}\tag{1}$$ 
One may think of this as having a UNION of $a$ intervals $\left[\frac{ab-1}{cb-1},\frac{ab-1}{(c-1)b+1}\right]$  on the real line and proving that, for some $a$, these can trap any finite discrete set of rational numbers having cardinality $b-2$.
For example, Let's assume $b=10$, $x=\{\frac{12}{5},\frac{17}{5},\frac{59}{10},\frac{187}{20},15,21,\frac{67}{2},39 \}$, then $(1)$ is true for $a=4$ and $a=5$ and therefore the intervals for $a=4,5$ contain or "trap" all the members of set $x$. In the diagram, the red/green regions are intervals $\left[\frac{ab-1}{cb-1},\frac{ab-1}{(c-1)b+1}\right]$ for all $c<a$.

What mathematical tools can one employ to tackle this?
WHAT I DID:
I solved the inequality $(1)$ for $a$ and got
$$\large x_i\left(1-\frac{2}{b}\right)+\frac{x_i+1}{b}+cx_i\geq a\geq \frac{x_i+1}{b}+cx_i\tag{2}$$
Note that $\frac{x_i+1}{b}+cx_i$ is common on both sides of the inequality.
Now I'll just have to plug in the values of $x_i$ to get $b-2$ inequalities and check for a common value of $a$ for any $c$. 
For example, Let $x_i \in \{  \frac{101}{4},\frac{3001}{7} \}$ and $b=4$
then we have inequalities 
$\frac{307}{16}+\left(\frac{101}{4}\right)c_1\geq a\geq \frac{105}{16}+\left(\frac{101}{4}\right)c_1$
$\frac{911}{32}+\left(\frac{301}{8}\right)c_2\geq a\geq \frac{309}{32}+\left(\frac{301}{8}\right)c_2$
Put $c=2$ in the first inequality and $c=1$ in the second inequality to get a common $a=58$.
But how should one go about to prove that there will always be some value of $a$ which satisfies the above set of inequalities? I calculated numerically for some random $b$ and $x_i$ and always managed to get some $a$, i.e., there existed some $a$ which trapped random $x_i$ given some random $b$. A counterexample may prove the negative but I was unable to find any.
 A: Here's what I did. I didn't get too far, but perhaps this will be of help.
First, since all of the relevant quantities are non-zero, let us define $y_i\triangleq 1/x_i$ and invert the inequality:
$$\frac{(c_i-1)b+1}{ab-1} \leq y_i \leq \frac{c_ib-1}{ab-1}$$
Note that each $y_i$ is a rational number between $0$ and $1$, exclusive. So define $y_i=p_i/q$, where $q$ is a common denominator across all $i=1,2,\dots,b-2$, and each $p_i\in\mathbb{N}$  is between $1$ and $q-1$, inclusive. So we have
$$\frac{(c_i-1)b+1}{ab-1} \leq \frac{p_i}{q} \leq \frac{c_ib-1}{ab-1}$$
$$q((c_i-1)b+1) \leq p_i(ab-1) \leq q(c_ib-1)$$
$$q(1-b) \leq p_i(ab-1) - qc_ib \leq -q$$
$$q(1-b/2) \leq p_i(ab-1) - qc_ib + qb/2 \leq q(b/2-1)$$
$$\left| 2(ab-1)p_i - 2qb(c_i-1/2) \right| \leq q(b-2)$$
Here's the geometrical interpretation of this expression. The quantity $2qb(c_i-1/2)$ represents a finite grid $$\{qb,3qb,5qb,..,2qba-qb\}.$$ The quantity $2(ab-1)p_i$, on the other hand, represents the $b-2$ chosen points, which must lie on a different finite grid
$$\{2(ab-1),4(ab-1),\dots,2(q-1)(ab-1)\}.$$
So the challenge is to find a way to ensure that $b-2$ points selected from this second grid lie within $q(b-2)$ of the points from the first grid. Note that this is not trivially satisfied because this distance $q(b-2)$ is strictly less than half the spacing between the $c_i$ grid points. So there are indeed spaces that must be avoided when choosing $a$.
