Andaleeb sent in this excellent solution.
The diagram shows some of the vertical lines drawn for values of x
between 0 and 1 as described in the question. The lines are of
height 1 unit at x = 0 and 1, of height ${1\over 2}$ units at $x=
{1\over 2}$ , of height ${1\over 4}$ units at $x= {k\over 4}$ and
${1\over 8}$ units at $x= {k\over 8}$ and so on... up to ${1\over
2^5}$ at ${k\over 2^5}$ where $k$ is a positive integer.
n 
$0$ 
$1$ 
$2$ 
$3$ 
$4$ 
$5$ 
$6$ 
$\dots$ 
$n$ 
$n+1$ 
Height 
$1$ 
${1\over 2}$ 
${1\over 4 }$ 
${1\over 8}$ 
${1\over 16}$ 
${1\over 32 }$ 
${1\over 64}$ 
$\dots$ 
${1\over 2^n}$ 
${1\over 2^{n+1`}}$ 
Lines cut 
$2$ 
$1$ 
$2$ 
$4$ 
$8$ 
$16$ 
$32$ 
$\dots$

$2^{n1}$ 
$ 2^n$ 
Thus if the height $h$ lies in ${1\over 2^n} > h> {1\over
2^{n+1}}$ then the number of lines cut is given by $$ 2 + 1 + 2 + 4
+ 8 + ... + 2^{n1} = 2 + {{2^n  1} \over {2  1}} = 2^n + 1.$$ As
$n$ tends to infinity the height of the lines tends to 0 and the
number of lines cut tends to infinity.