You can use that a set function $\mu$ defined on a measurable space such that $\mu(\emptyset)=0$, $\mu$ is finitely additive and is continuous from below is a measure on that space.
Also, you can define a double sequence $(s_{n,m})_{n,m\geq 1}$ given by $s_{n,m}=\sum_{k=1}^m \mu_n(E_k)=\mu_n(\bigcup_{k=1}^m E_k)$ and then show that
$$\lim\limits_{n\to\infty}\lim\limits_{m\to\infty} s_{n,m}=\sup\limits_n\sup\limits_m(s_{n,m})=\sup\limits_m\sup\limits_n(s_{n,m})=\lim\limits_{m\to\infty}\lim\limits_{n\to\infty} s_{n,k}$$
Note that the equalities between the supremum and the limit is because in each index, the sequence $(s_{n,k})$ is monotonically increasing.
Since $\mu_n \leqslant \mu$ for all $n$, we have
$$\int f\,d\mu_n \leqslant \int f\,d\mu$$
for all $n$, and therefore
$$\lim_{n\to\infty} \int f\,d\mu_n = \sup_n \int f\,d\mu_n \leqslant \int f\,d\mu.$$
Conversely, we have
$$\int f\,d\mu = \sup \Biggl\{ \int s\,d\mu : 0 \leqslant s \leqslant f,\; s\text{ simple}\Biggr\}.$$
So fix an arbitrary simple $s$ with $0 \leqslant s \leqslant f$. By the case for simple functions,
$$\int s\,d\mu = \lim_{n\to\infty} \int s\,d\mu_n \leqslant \lim_{n\to\infty} \int f\,d\mu_n.$$
Taking the supremum over all admissible $s$ gives
$$\int f\,d\mu \leqslant \lim_{n\to\infty} \int f\,d\mu_n,$$
so together with the first part, we have the monotone convergence theorem, supposing your argument for simple functions is correct. (It should be, that case is simple.)
Best Answer
Hint
$$\lim_{n \to \infty} \sum_{i=1}^{\infty} \mu_n(u_i) = \sup_{n \in \mathbb{N}} \sup_{k \in \mathbb{N}} \sum_{i=1}^k \mu_n(u_i) = \sup_{k \in \mathbb{N}} \sup_{n \in \mathbb{N}} \sum_{i=1}^k \mu_n(u_i) = \ldots$$
The first equality follows from the fact that the sequence of measures is increasing (i.e. $\lim = \sup$).