I proved the above problem as follows but received feedback that I'm not certain I understand. Can someone help me determine where I went wrong in the proof?
Let A⊆R be a set. Suppose M∈R where M>0 and |x|≤M for all x∈A. Thus -M≤x≤M for all x∈A. It is reasonable to conclude that A is bounded with an upper bound M and a lower bound -M.
Let A be bounded. It follows that there exists a lower bound a∈R and an upper bound b∈R. Thus |a|≤x≤|b| for all x∈A.
Let |a|≤|b|. Then -|b|≤-|a| and -|b|≤x≤|b| for all x∈A. It follows that -(|b|+1)0 and |x|≤M for all x∈A.
Let |b|<|a|. Then -|a|≤x<|a| for all x∈A. It follows that -(|a|+1)0 and |x|≤M for all x∈A.
I thought I did a pretty good job but my professor thought my two cases should be different. She also suggested that I missed a step, that I need to assume that |a|≤|b| means that x ≤-a =|a|≤|b| but I'm really not following that suggestion at all.
This is incorrect; you put absolute values where they don't belong. The definition of upper and lower bounds tells you that $a\le x\le b $ for all $x \in A$.
This looks like a broken formula.
This is a misuse of "let". It is meant to signal the reader that some assignment has been made, like "let $c=4$". Here, $a$ and $b$ are already introduced, you can't apply "let" to them as if they were fresh variables. It's better to write: "There are two cases. Case 1: $|a|\le |b|$. Case 2: $|a|>|b|$."
One problem with the two final paragraphs is that you never said what $M$ is. If you want to convince the reader that $M$ exists, it helps to spell out what it is: $M=|b|$ in the first case, for example.
(You could shorten the proof with $M=\max(|a|,|b|)$, unifying the cases.)