How can the VNM completeness axiom be derived from the transitivity and continuity axioms?
I think I just found a counterexample to your teacher's claim.

Suppose completeness is violated completely (sorry for the pun): $L\not\succsim L'$ and $L'\not\succsim L$ for any $L,L'\in\mathcal L$, i.e. no two lotteries are comparable. 

Then I claim that transitivity and continuity are both satisfied vacuously. 

Transitivity: $L\succsim L'$ and $L'\succsim L''$ imply $L\succsim L''$ for any $L,L',L''\in\mathcal L$. But since no two lotteries are comparable by $\succsim$, the antecedent is always false, and so transitivity is always satisfied. 

Continuity: for any $L,L',L''\in\mathcal L$, the sets $\{\alpha\in[0,1]:\alpha L+(1-\alpha)L'\succsim L''\}$ and $\{\alpha\in[0,1]:L''\succsim\alpha L+(1-\alpha)L'\}$ are both closed. Again, since no two lotteries are comparable, both sets are empty and therefore closed. Hence, continuity is satisfied in a vacuous sense.