Let $(X, \mathcal A, \mu)$ be a $\sigma$-finite measure space. Let $\mathcal F$ be the collection of all measurable sets with finite measure. We define a pseudometric metric $d_\mu$ on $\mathcal F$ by $d_{\mu}(A, B) := \mu(A \triangle B)$. Then $d_\mu$ becomes a metric when $\mathcal F$ is considered modulo the equivalence relation $\sim$ defined by $$ A \sim B \iff \mu(A \triangle B) = 0 \quad \forall A,B \in \mathcal F. $$
I'm trying to generalize this result from finite measure space to $\sigma$-finite measure space, i.e.,
Theorem: If the $\sigma$-algebra $\mathcal A$ is countably generated, then $d_\mu$ is separable.
Could you have a check on my proof?
My attempt: By $\sigma$-finiteness assumption, there is an increasing (w.r.t. inclusion) sequence $(X_n) \subset \mathcal A$ such that $\bigcup_n X_n = X$ and $\mu(X_n) < \infty$. Let $$ \mathcal A_n := \{X_n \cap A \mid A \in \mathcal A\} \subset \mathcal F \subset \mathcal A \quad \forall n \in \mathbb N^*. $$
Because $\mathcal A$ is countably generated, so is $\mathcal A_n$. Let $d_{n}$ be the restriction of $d_\mu$ to $\mathcal A_n$. By this result, $d_n$ is separable. Let $\mathcal D_n$ be a countable dense subset of $\mathcal A_n$. Let $\mathcal D := \bigcup_n \mathcal D_n$. Then $\mathcal D$ is countable. Let's prove that $\mathcal D$ is dense in $\mathcal F$.
Fix $\varepsilon>0$ and $A \in \mathcal F$. Let $A_n := X_n \cap A \in \mathcal A_n$. Then $A_n \nearrow A$. By continuity of measure from below, there is $N$ such that $$ d_\mu(A_N, A) = \mu(A \setminus A_N) < \varepsilon. $$
There is $D \in \mathcal D_N$ such that $d_N (D, A_N) < \varepsilon$. It follows that $$ d_\mu( D, A) \le d_\mu (D, A_n) + d_\mu(A_N, A) = d_N (D, A_n) + d_\mu(A_N, A) < 2 \varepsilon. $$
This completes the proof.