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 $\mathcal{M} \supseteq \sigma(\mathcal{F})$.
Your proof that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)$ is fine.
The proof that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$ is OK, but it can be done in a simpler way.
Proof that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$ :
Since $S_2 \subset \mathcal{O}$, it follows immediately that $$\sigma(S_2) \subseteq \sigma(\mathcal{O}) = \mathcal{B}(\mathbb{R}^n)$$
Since, any open set can be written as a countable union of rectangles, we have that
$ \mathcal{O} \subseteq \sigma(S_2)$. So, we have
$$\mathcal{B}(\mathbb{R}^n)= \sigma( \mathcal{O}) \subseteq \sigma(S_2)$$
So, we have $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$.
Best Answer
Since the title is Borel $\sigma-$ algebra, I am pretty sure $\tau$ is usual metric topology.
Then $\sigma(\tau)$ is uaual Boral $\sigma$ algebra.
I will show that $\sigma(\tau)\subseteq \sigma(\mathcal{O}_2)$ and it suffices to prove that $\sigma(\mathcal{O}_1)\subseteq \sigma(\mathcal{O}_2)$ since you already prove that $\sigma(\mathcal{O}_1)=\sigma(\tau)$.
So our goal is to prove $\mathcal{O}_1\subseteq \sigma(\mathcal{O}_2)$.
Let $(a_1,b_1)\times \cdots \times (a_k,b_k)\in \mathcal{O}_1$ be given.
And note that $$(a_1,b_1)\times \cdots \times (a_k,b_k)=\bigcup_{n=1}^{\infty} [(-\infty,b_1)\setminus \left(-\infty,a_1+\frac{1}{n}\right)]\times \cdots \times \bigcup_{n=1}^{\infty} [(-\infty,b_k)\setminus \left(-\infty,a_k+\frac{1}{n}\right)] . $$
Clearly, right hand side is an element of $\sigma(\mathcal{O}_2)$ since it is closed under set subtraction and countable union and each $(-\infty,b_i)$ and $(-\infty,a_i+n^{-1})$ are in $\mathcal{O}_2$.
Therefore, $\mathcal{O}_1\subseteq \sigma (\mathcal{O}_2)$.
Since $\sigma(\mathcal{O}_2)$ is a sigma algebra containing $\mathcal{O}_1$ and $\sigma(\mathcal{O}_1)$ is the smallest sigma algebra containing $\mathcal{O}_1$, we have $\sigma(\tau)=\sigma(\mathcal{O}_1)\subseteq \sigma(\mathcal{O}_2)$.
We are done.