How to simulate a die given a fair coin
What you can do, is to employ a method called rejection sampling: 


Flip the coin 3 times and interpret each flip as a bit (0 or 1).
Concatenate the 3 bits, giving a binary number in $[0,7]$.
If the number is in $[1,6]$, take it as a die roll.
Otherwise, i.e. if the result is $0$ or $7$, repeat the flips.


Since $\frac 68$ of the possible outcomes lead to termination in each set, the probability of needing more than $l$ sets of flips to get a die roll is $(1-\frac 68)^l = \dfrac 1{4^l}$. Hence, this method is efficient in practice.

Improvements:

@Angel's answer points out, that the number of coin flips in each set but the first can be reduced from 3 to 2, by using the distinction between a $0$ and a $7$ as the first bit for the next set.

@Emanuele Paolini explains, how you can reduce the number of rerolls, if you need multiple die rolls.