I know that is has to be bounded because its finite, but I'm guessing there is some sort of mathematical proof for this?
Bounded in what sense? That the diameter of the set is finite? This would follow since $\displaystyle \Omega=\left\{|a_k-a_j|:a_k,a_j\in S,\text{ }j\ne k\right\}$ is finite and each element is a finite number and thus $\displaystyle \text{diam }S=\max\text{ }\Omega<\infty$
What Drexel28 is trying to say is...
Assume that $\displaystyle S$ is finite but not bounded above.
Set $\displaystyle max\{S\}=m_0,$ but then the set is bounded by $\displaystyle m_0.$
Therefore, there must be some $\displaystyle m_1\in S: m_1 > m_0,$ but then the set is bounded by $\displaystyle m_1$... ad infinitum.
The set must be infinite to be unbounded.
A similar argument shows the set is bounded below...