Uniform Integrability: domination implies UI
Solution 1:
It might work to consider that $\sup_n |f_n|$ itself is measurable and bounded by $g$, so it's integrable; and $$\lim_{a \rightarrow \infty} \int_{|f_k| \ge a} \sup_n |f_n| \, \mathrm{d}\mu = 0 \; \; \mathrm{for} \; \mathrm{every} \; k$$ by the argument you gave above. This should be enough.