A set contains $0$, $1$, and all averages of any subset of its element, prove it contains all rational number between 0 and 1 Assume we have a set $S$ $\subseteq \mathbb{R}$, we know $0,1 \in S$. Also, $S$ has the property that if $A \subseteq S$ is finite, the average of elements in $A$ is in $S$. We want to prove that any rational number between 0 and 1 is contained in $S$.
Can someone please give me a hint?  I tried to use induction but it seems hopeless to me...

Here is a more explicit construction than given in other answers.
As other answers, first notice that it’s quite easy to get all dyadic numbers in $[0,1]$, i.e. numbers of the form $\frac{n}{2^a}$, where $0 \leq n \leq 2^a$.
Now suppose you want to get $\frac{p}{q}$.  If you can find $q$ distinct dyadic numbers $x_1$, …, $x_q$ whose sum is exactly $p$, then the average of these will be $\frac{p}{q}$.  How can you find those distinct dyadic numbers with the right sum?  There are lots of ways to do this, but here’s one.  Since the dyadic numbers are dense, you can take them to be close approximations of $\frac{p}{q}$ itself.  First pick a denominator $2^a$ to use (for all the $x_i$).  Now, using this denominator, take all but one of the numbers $x_i$ to be below $\frac{p}{q}$, but as close as possible to it; and then pick the last number,$x_q$, to be whatever is needed to balance out the total shortfall of the rest, to make the total sum $p$.
We’re nearly done, but not quite.  Let’s look at what this method gives for making 1/3.  We’re looking for three dyadics with sum 1.  With denominator 4, we get 0, 1/4 as the nearest ones below 1/3, and then 3/4 to bring the total up to 1.  With denominator 8, we get 1/8, 2/8 as the ones below 3, and then 5/8 as the last number.  With denominator 16, we get 4/16, 5/16, and 7/16.
What about for 1/5?  We want five dyadics summing to 1, starting with four below 1/5.  Trying denominator 4, the four closest aproximations by quarters below 1/5 are… 0, –1/4, –2/4, and –3/4.  Oops.  By eighths, not much better.  By sixteenths, we have 3/16, 2/16, 1/16, and 0/16; and then 10/16 is just right to bring the sum up to 1.  Success!
What about for 2/3?  We want three dyadics with sum 2.  Trying quarters, we get 2/4 and 1/4 as the approximations below it… but then we would need 5/4 to bring the sum up to 2.  Trying with eighths, we get 5/8 and 4/8 as the approximations-below; and 7/8 to finish.  Success!
Generally, for $\frac{p}{q}$: trying this method, with denominator $2^a$, we get approximations-below of $k/2^a$, $(k-1)/2^a$, …, $(k–q+2)/2^a$, where $k = \lfloor{2^a p / q}\rfloor$, i.e. the largest numerator such that $k/2^a \leq p/q$.  Then the last number we pick must bring the sum of the numerators up to $2^a p$, so must be $(2^a p – \sum_{i=0}^{q-2}(k-i))/2^a$.
This will always give a set of $q$ distinct dyadic numbers with average $p/q$.  As the examples above show, sometimes some of these dyadics will be outside $[0,1]$.  But if you take $a$ large enough (how large? I’ll leave that as an exercise…) then these numbers will be in $[0,1]$, and so you’re done.
