[oldmathguy]

Here's a practical way to view a counterexample. It's a problem about
slopes of pieces of the graph since (f(x)-f(y)<E(x-y) means that the
average rate of change (f(x)-f(y))/(x-y) must be <E. So do this:
Draw a jagged continuous graph f(x)which is piecewise-linear ( made up of
line segments) with infinitely many larger and larger slopes. If you can imagine
this, mark option III false and go on to the next problem. If you want a rigorous
definition of a function try this:

f(0)=f(1) = 0,
between 0 and 1/2 it has slope 1 going up to (1/4,1/4) and slope -1 down
to (1/2, 0),
from x=1/2 to x=3/4, it goes up to height 2/8 with slope 2 to the half-way
point at x=5/8 and down with slope -2 to (3/4,0)
Continue in this way, with slopes n and -n between points (1- 1/2^{n-1}, 0)
and (1 - 1/2^n, 0), achieving maximal height n/2^{n+1} halfway
between these two points.
Notice that the limit of n/2^{n+1} as n--> infinity is 0, so the function
is continuous at x=1.

Then given any E there is an integer n>E and a piece of the graph with
slope n. So it won't be true that the slope( f(x)-f(y))/(x-y) < E for all
x and y in [0, 1].