Darboux Integrability epsilon-delta proof

Hello all, I'm having trouble proving this theorem.

Suppose f is Darboux integrable, then for all $\displaystyle \epsilon > 0$ there exists a $\displaystyle \delta > 0$ such that mesh (P) < $\displaystyle \delta$ implies U_p(f) - L_p(f) < $\displaystyle \epsilon$.

Proof:

Let $\displaystyle \epsilon > 0 $, and suppose f is Darboux Integrable.

If f is a constant we are done, if not pick $\displaystyle mesh (P) < \frac{\epsilon}{\sum\limits_{k=1}^n M_k - m_k}$, where k is the nth sub-interval of the partition P. M_k is the supremum of the function in that respective sub-interval, m_k is the infimum.

It follows that,

$\displaystyle \vert U(f) - L(f) \vert < \vert U_p (f) - L_p (f) \vert < \epsilon $.

However, my instructor said that, I cannot pick that delta because the suprema and infima of the sub-intervals depend on the partition. I left out some steps, because the problem with the proof is the choice of delta.

Any thoughts on this? Thank you