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 $\frac{1}{2}$ units at $x = \frac{1}{2}$ , of height $\frac{1}{4}$ units at $x = \frac{k}{4}$ and $\frac{1}{8}$ units at $x = \frac{k}{8}$ and so on... up to $\frac{1}{2^{5}}$ at $\frac{k}{2^{5}}$ where $k$ is a positive integer.

n 0 1 2 3 4 5 6 ... n n+1

Height 1 $\frac{1}{2}$ $\frac{1}{4}$ $\frac{1}{8}$ $\frac{1}{16}$ $\frac{1}{32}$ $\frac{1}{64}$ ... $\frac{1}{2^{n}}$ $\frac{1}{2^{n + 1 `}}$

Lines cut 2 1 2 4 8 16 32 ... $2^{n - 1}$ $2^{n}$

Thus if the height

h

lies in

\frac{1}{2^{n}} > h > \frac{1}{2^{n + 1}}

then the number of lines cut is given by

2 + 1 + 2 + 4 + 8 + . . . + 2^{n - 1} = 2 + \frac{2^{n} - 1}{2 - 1} = 2^{n} + 1 .

n

tends to infinity the height of the lines tends to 0 and the number of lines cut tends to infinity.

n	0	1	2	3	4	5	6	...	n	n+1
Height	1	$\frac{1}{2}$	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{16}$	$\frac{1}{32}$	$\frac{1}{64}$	...	$\frac{1}{2^{n}}$	$\frac{1}{2^{n + 1 `}}$
Lines cut	2	1	2	4	8	16	32	...	$2^{n - 1}$	$2^{n}$