Use pigeonhole on the intervals $[n^2,n^2+2n]$ for $n=1,\cdots,9$, and the set $\{100\}$. This is also true if we consider $A=\{1,2,\cdots,120\}$.

Look at the sets $\{1 \}, \{2,3,4 \}, \{5,6,7,8,9 \},\dots,\{82,83,\dots,100 \}$. Those are exactly $10$ sets, by the pigeonhole principle there will be at least two elements of one of those sets. Choose those and we're done. Edit: rip flake 2 fast Edit2: nvm we have different proofs

Consider $A_{n}=[n^2,(n+1)^2)(1\leq n\leq 9)$ and $\{100\}$.By PHP, we must choose $2$ numbers from some $A_{i}$.This $2$ numbers satisfy the condition.