$m^*(E_n)=\inf{\left\{\displaystyle\sum_{k=0}^\infty \ell(I_k)\mid E_n\subseteq\bigcup_{i=0}^\infty I_k \right\}}$, where $I_k$ is an open bounded interval. Since what is on the left hand side of your inequality is one of these possible coverings, the infimum, $m^*(E_n)$ is less than or equal to that. This means you can find an $\epsilon>0$ such that $m^*(E_n) + \epsilon > \displaystyle\sum_{k=0}^\infty \ell(I_k)$, where $\{I_k\}_{k=0}^\infty$ is one possible colleciton of bounded open sets containing $E_n$. Since we have a countable collection of these sets, $E_n$, we can associate to each set a natural number $n$ so that your inequality holds for the choice of $\dfrac{\epsilon}{2^n}$ for the $n$th set.
Your proof is almost correct, but it is not OK.
You defined
$\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) | \mu_1(A)=\mu_2(A) \rbrace$.
Then you take $\lbrace A_n \rbrace$ be a sequence of non-decreasing sets from $\mathcal{M}$. But when you "disjointify" the sets, taking
$B_k=A_k- \bigcup_{j=1}^{k-1}A_j$, all you can say is that $B_i$ is in $\sigma(\mathcal{F})$. You can not say $B_i$ is in $\mathcal{F}$ nor $B_i$ is in $\mathcal{M}$.
The way to correct your proof is simple. Here it is:
Let $\mu_1$ and $\mu_2$ be two finite measures defined on $\sigma(\mathcal{F})$ such that, $\forall A \in \mathcal{F}$, $\mu_1(A)=\mu_2(A)$. Show that they must agree on $\sigma(\mathcal{F})$.
Let $\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) \,| \, \mu_1(A)=\mu_2(A) \rbrace$. Let us show that $\mathcal{M}$ is a monotone class.
First, let $\lbrace A_n \rbrace$ be a monotone non-decreasing sequence of sets from $\mathcal{M}$.
So, for all $n$, $\mu_1(A_n) =\mu_2(A_n)$.
Since $\lbrace A_n \rbrace$ is be a sequence of non-decreasing sets in $ \sigma(\mathcal{F})$, for any measure $\nu$ defined on $ \sigma(\mathcal{F})$, we have
$$\nu\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\nu(A_n)$$
So applying this to $\mu_1$ and $\mu_2$, we get
$$\mu_1\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\mu_1(A_n)=\lim_{n \to \infty}\mu_2(A_n) =\mu_2\left (\bigcup\limits_{n=1}^{\infty}A_n \right) $$
So $\bigcup\limits_{n=1}^{\infty}A_n \in \mathcal{M}$, and we conclude that $\mathcal{M}$ is closed under monotone non-decreasing union.
Now, to show $\mathcal{M}$ is closed under decreasing limits of sets.
Let $\lbrace C_n \rbrace$ be a monotone non-increasing sequence of sets from $\mathcal{M}.$
So, for all $n$, $\mu_1(C_n) =\mu_2(C_n)$.
Using continuity from above of a finite measure, we have
$$\mu_1\left (\bigcap\limits_{n=1}^{\infty}C_n \right) = \lim_{n \to \infty}\mu_1(C_n)=\lim_{n \to \infty}\mu_2(C_n) =\mu_2\left (\bigcap\limits_{n=1}^{\infty}C_n \right) $$
So $\bigcap\limits_{n=1}^{\infty}C_n \in \mathcal{M}$, and we conclude that $\mathcal{M}$ is closed under monotone non-increasing intersection.
Thus, $\mathcal{M}$ is a monotone class containing $\mathcal{F}$ and by monotone class theorem, $\mathcal{M} \supseteq \sigma(\mathcal{F})$. So, for all $A\in\sigma(\mathcal{F})$, we have $\mu_1(A)=\mu_2(A)$.
Remark: As defined, $\mathcal{M}$ may be bigger than $\sigma(\mathcal{F})$, but all we need to prove the result is $\mathcal{M} \supseteq \sigma(\mathcal{F})$.
Best Answer
Given a union of sets $\bigcup_{n = 1}^\infty F_n$, you can create a disjoint union of sets as follows.
Set $G_1 = F_1$, $G_2 = F_2 \setminus F_1$, $G_3 = F_3 \setminus (F_1 \cup F_2)$, and so on. Can you see what $G_n$ needs to be?
Using $m(\bigcup_{n = 1}^\infty G_n)$ and monotonicity, you can prove $m(\bigcup_{n = 1}^\infty F_n) \leq \sum_{n = 1}^\infty m(F_n)$.