How can I prove $2\sup(S) = \sup (2S)$? 
Let $S$ be a nonempty bounded subset of $\mathbb{R}$ and $T = \{2s : s \in S \}$. Show $\sup T = 2\sup S$

Proof
Consider $2s  = s + s \leq \sup S + \sup S = 2\sup S $. $T \subset S$ where T is also bounded, so applying the lub property, we must have $\sup T \leq 2 \sup S$.

On the other hand $2s + s - s \leq \sup T + \sup S - 3\sup S \implies 2\sup S \leq 2s + 2\sup S \leq \sup T $. Which gives the desired result.

Okay I am really worried about my other direction. Especially  $2\sup S \leq 2s + 2\sup S$, do I know that $2s$ is positive? 
Also in the beginning, how do I know that $\sup S \leq 2 \sup S$? How do I know that the supremum is positive?

You can't assume $s$ is positive, nor can you assume $\sup S$ is positive.
Your proof also assumes a couple of other weird things:


*

*$T \subset S$ is usually not true.

*$2s + s - s \le \sup T + \sup S - 3\sup S$ is not necessarily true.  Why would $-s \le -3\sup S$?


The first part of your proof is actually correct, ignoring the $T \subset S$ statement.
What you are saying is that any element of $T$, say the element $2s$, is bounded above by $2 \sup S$; thus $2 \sup S$ is an upper bound on $T$; thus $2 \sup S \ge \sup T$ by the least upper bound property.
For the second part of the proof, you need to show that $\sup T \ge 2 \sup S$.
To do this, you need to show that $\frac{\sup T}{2}$ is an upper bound on $S$.
This will imply $\frac{\sup T}{2} \ge \sup S$ by least upper bound property.
