Prove that $[1,2]+[-2,-1]=[-1,1]$ $A+B=\{x\in\mathbb{R}: x=a+b$ for some $a\in$ A and $b\in B\}$ A=[1,2], B=[-2,-1], prove A+B=[-1,1] 

This is my attempt but my professor said it's not good enough. 
Proof: since all elements $x\in A+B$ are defined to be $x=a+b$ for some $a\in A, b\in B$, and since the smallest element in A is 1, and that of B is -2, the smallest element $x\in A+B$ is therefore -1. By the same logic, the largest element $x\in A+B$ is 1. Since $\forall x\in A+B$ must be greater than or equal to its smallest element and smaller than or equal to its biggest element, we have $-1 \leq x \leq 1, \forall x\in A+B$. Therefore $A+B=[-1,1]$
My professor said that this only proves $A+B\subseteq [-1,1]$. Comments?
 A: In proving equality of sets $Q=W$, one usually would prove two inclusions $Q\subseteq W$ and $W\subseteq Q$.
So let $x\in A+B$ be given an prove that it belongs to $[-1,1]$. Thus $A+B\subseteq [-1,1]$. After that assume $x\in[-1,1]$ and prove that it belongs to $A+B$ by writing it as a sum of the form $x=a+b$ with $a\in A$ and $b\in B$.

The part you have already proven is that $x\in A+B$ has $a\geq 1$ and $b\geq -2$ so $x\geq 1-2=-1$ and similarly $x\leq 2-1=1$. Hence $x\in[-1,1]$. Conclusion is that $A+B\subseteq [-1,1]$.
For the other part, assume first that $x\in[-1,0]$. Then $-1\leq x=1+b\leq 0$ implies $-2\leq b\leq -1$. So in this case $x=a+b$ with $a=1$ and $b\in B$.
Then assume $x\in[0,1]$ and write $0\leq x=2+b\leq 1$. Then again $b\in[-2,-1]=B$ so we have written $x=a+b$ with $a=2$ and $b\in B$.
To sum up, for $x\in[-1,0]\cup[0,1]=[-1,1]$ we can write $x=a+b$ for some $a\in A$ and some $b\in B$. Thus $[-1,1]\subseteq A+B$ and we are done.

Perhaps you can generalise this result yourself an go on to prove that $[a,b]+[c,d]=[a+c,b+d]$.
A: Generally,I don't think you really need a difficult proof.you are setting "data" at each of the number equation.This means that [1,2]+[-2,-1]=[-1,1].Putting in algebra,
[First Data A,First Data B]+[Second Data A,Second Data B]=[Total Data A,Total Data B]
Or like string said, [a,b]+[c,d]=[a+c,b+d]
I disagree of a proof required for this because its like trying to proof 1+2=3 while having 2+4=6 putting at a box/data.
