Real analysis supremum proof 
Let $A$ be a non-empty bounded sub-set of $\mathbb{R}$. Let
  $B\subset\mathbb{R}$, given by $$B=\left\{\frac{a_1+2a_2}{2} \,\Bigg|\,a_1,a_2\in
 A\right\}$$ Express $\sup B$ in terms of $\sup A$.

My attempt:
Suppose $a_1,a_2\in A$ and $b\in B$. 
Then $a_1 \leq \sup (A)$ and $a_2\leq \sup (A)$
So $a_1 + 2a_2\leq 3sup (A)$.
This gives $\frac{a_1 + 2a_2}{2}\leq \frac{3\sup (A)}{2}$.
This means that $\frac{3\sup (A)}{2}$ is an upperbound and $\sup(B)\leq\frac{3\sup (A)}{2}$. 
Now let $\epsilon>0$. 
$a_1>\sup(A) - \frac{2\epsilon}{3}$
$a_2>\sup(A) - \frac{\epsilon}{3}$
This gives $\frac{a_1 + 2a_2}{2}> \frac{3\sup (A)}{2}-\epsilon$. 
So this means that $\frac{3\sup(A)}{2}\leq\sup(B)$
So $\sup(B)=\frac{3\sup(A)}{2}$.
Is this correct? And how can I improve my proofs?

Your proof is correct.
You can also do this: let $(a_n)_n$ be a sequence in $A$ such that $a_n \to \sup A$. Then $$B \ni \frac{a_n + 2a_n}2 \to \frac{\sup A + 2\sup A}2 = \frac32 \sup A$$ so $\sup B \ge \frac32 \sup A$.
