Assume there was an irrational number in $ S$, only finitely much, and number them $ x_{1}\le\ldots\le x_{n}$. Now $ x_{1}=\frac{a+x_{k}}{2}$ with $ a<x_{1}$ rational. So, we can express $ x_{1}$ as a (nonnegative) linear combination of rationals and irrationals with higher indices (over $ \mathbb{Q}$). Now let $ t<n$ be the smallest index for which this is impossible, then $ x_{t}=\frac{a+b}{2}$ with $ a>x_{t}>b$ and necessarily $ b$ irrational (else $ a$ irrational but must have higher index). But then $ b\in\{x_{1},...,x_{t-1}$, which is a nonnegative linear combination of $ x_{t},...,x_{n}$. Now, if the coefficient in that combination of $ x_{t}$ is not $ 1$, then we can solve it for $ x_{t}$ as a (nonnegative) linear combination of $ x_{t+1},\ldots,x_{n}$, contradiction. So $ x_{t}$ appears in coefficient $ 1$ and, by nonnegativity, the rest appears in coefficient $ 0$, so $ b=x_{t}$, contradiction. So $ t=n$ and thus every $ x_{i}$ is a rational multiple of $ x_{n}$. But then $ x_{n}=\frac{a+b}{2}$ with $ b>x_{n}$ rational and $ a<x_{n}$ irratiional. But then $ a=qx_{n}$, and thus $ b=(2-q)x_{n}$ would be irrational, contradiction. So there can be no irrational number in $ S$.

Another solution: We consider $\mathbb R$ as a vectorial space over $\mathbb Q$. Let $V$ be the subspace generated by $S$. If $V\neq\mathbb Q$, take $v_1,v_2,\ldots$ in $S$ such that $\{1,v_1,v_2,\ldots\}$ is a basis for $V$. Each $x\in S$ can be written as $a_0+a_1v_1+...$, with $a_i\in\mathbb Q$. Consider the number $x$ whose coefficient of $v_1$ is maximal. It is the average o two other numbers. Clearly, both of them have the same $v_1$ coefficient as $x$. We find a contradiction by choosing $x$ as large as possible.$\square$