Let $S=\{a_1,\dots,a_n\}$ with $a_1<\cdots<a_n$. Consider the function, $\phi(x,y)=x+y\cdot a_n$. I claim $\phi : S\times S \to \mathbb{Z}^+$ is injective, which will then yield the desired result. Suppose $(x,y),(x',y')\in S\times S$ with $\phi(x,y)=\phi(x',y')$. Notice, if $x=x'$ or $y=y'$, we directly have $(x,y)=(x',y')$, so suppose $x\ne x'$ and $y\ne y'$. Now, $x+y a_n = x'+y' a_n\Rightarrow |x-x'| = a_n |y-y'|\ge a_n$, as $y\ne y'$. But since $x,x' \in[a_1,a_n]\cap \mathbb{Z}$ with $a_1>0$, $|x-x'|<a_n$. This contradiction shows $(x,x')=(y,y')$; thus $\phi$ is injective on $S\times S$, which has cardinality $n^2$, thus concluding the proof.