A diagonalization argument would work, we just need to take care how to approximate our functions. For ease of notation, define
$$\begin{align}H_1 &= F_1\\
H_{n+1} &= F_{n+1}\setminus F_n\end{align}$$
so that the $H_n$'s are pairwise disjoint and $\bigcup_{k=1}^n H_k = F_n$.
For all $n\geq 1$ we can take measurable simple functions $\{g^{(n)}_m\}_{m\geq 1}$ such that $g^{(n)}_m\mathop{\longrightarrow}_{m\to\infty} f^{(n)} := \chi_{H_n}f$. W.l.o.g. we can assume that $g^{(n)}_m$ is supported in $H_n$.
Define the functions
$$h_n := \sum_{k=1}^n g^{(k)}_n.$$
Clearly, for all $n\geq 1$ we have $\chi_{F_n}h_m\mathop{\longrightarrow}_{m\to\infty} \sum_{k=1}^n f^{(k)}=\chi_{F_n} f$ almost everywhere (since for every sufficiently large $m$ we have $\chi_{F_n}h_m = \sum_{k=1}^n g^{(k)}_m$), and since $\Omega=\bigcup F_n$ we're done.
It isn't true. Counterexamples include spaces like $\Omega = \{0,1\}^{[0,1]}$, a product of uncountably many copies of $\{0,1\}$ with the product topology. It is compact Hausdorff, hence $T_{3 \frac{1}{2}}$, and it is well known to be separable. But your $\sigma(C_b(\Omega))$ is the $\sigma$-algebra of Baire sets which here is strictly smaller than the Borel $\sigma$-algebra.
Let's represent elements of $\Omega$ as functions $\omega : [0,1] \to \{0,1\}$. The fact is that any Baire set in $\Omega$ can only depend on countably many coordinates, and in particular cannot be a singleton. To make this precise, for any countable $A \subset [0,1]$ let $\pi_A : \Omega \to \{0,1\}^A$ be the restriction map $\pi_A(\omega) = \omega|_A$. Then let $\mathcal{F}$ be the collection of subsets $E$ of $\Omega$ that are of the form $\pi_A^{-1}(F)$ for some countable $A \subset [0,1]$ and some $F \subset \{0,1\}^A$. Clearly $\Omega, \emptyset \in \mathcal{F}$, and $\mathcal{F}$ is closed under complements because $(\pi_A^{-1}(F))^c = \pi_A^{-1}(F^c)$. It is also closed under countable unions: suppose $E_1, E_2, \dots \in \mathcal{F}$, so that each $E_n = \pi_{A_n}^{-1}(F_n)$ for some countable $A_n \subset [0,1]$ and some $F_n \subset \{0,1\}^{A_n}$. Let $A = \bigcup_n A_n$ which is again countable, and let $\pi_n : \{0,1\}^A \to \{0,1\}^{A_n}$ be restriction, so that $\pi_{A_n} = \pi_n \circ \pi_A$. Then
$$\bigcup_n E_n = \bigcup_n \pi_A^{-1}(\pi_n^{-1}(F_n)) = \pi_A^{-1}\left(\bigcup_n \pi_n^{-1}(F_n)\right).$$
Hence $\mathcal{F}$ is a $\sigma$-algebra.
Consider the class $C_{\mathcal{F}}(\Omega)$ of continuous functions $f : \Omega \to \mathbb{R}$ which are $\mathcal{F}$-measurable. This is clearly an algebra containing the constants. It also separates points: for any $x \in [0,1]$ the function $\pi_x(\omega) = \omega(x)$ is continuous (by definition of the product topology) and $\mathcal{F}$-measurable, but if $\pi_x(\omega) = \pi_x(\omega')$ for all $x \in [0,1]$ then $\omega = \omega'$. Therefore, by Stone-Weierstrass, $C_\mathcal{F}(\Omega)$ is uniformly dense in $C(\Omega)$. That means every $f \in C(\Omega)$ is a pointwise limit of $\mathcal{F}$-measurable functions and is thus $\mathcal{F}$-measurable, so in fact $C(\Omega) = C_\mathcal{F}(\Omega)$.
Hence, $\sigma(C(\Omega)) \subset \mathcal{F}$.
On the other hand, for any fixed $\omega$, we have $\{\omega\} \notin \mathcal{F}$. For suppose $E \in \mathcal{F}$ with $\omega \in E$. Then $E = \pi_A^{-1}(F)$ for some countable $A$ and some $F \in \{0,1\}^A$. Since $A$ is countable, there exists $y \in [0,1] \setminus A$, so let
$$\omega'(x) = \begin{cases}
\omega(x), & x \ne y \\
1 - \omega(x), & x = y
\end{cases}$$
Since $y \notin A$, we have $\omega|_A = \omega'|_A$ and thus $\omega' \in E$. Therefore $E$ contains at least two points and cannot be the singleton $\{\omega\}$.
This is discussed a bit more in 6.10(i) of Bogachev's Measure Theory. He gives Engelking 2.7.12(c) as a reference for the theorem about uncountable products.
Best Answer
You mixed up indices when you wrote the inequality: $$ \int_{B_n} \phi_n \mathrm d\mu < M \mu(B_n) $$ You should use two different set of indices, one for the sequence $\phi_n$ and one for the sequence $B_n$. Let's use $m$ for the latter, then we have: $$ \int_{B_m} \phi_n \mathrm d\mu < M \mu(B_m) $$ Now, since $f$ can be unbounded on $B_m$, we must get clever. We have to slice up $B_m$ in a sequence $B_m^k$ of sets defined as $$ B_m^k = f^{-1}([0, k)) \cap B_m = \{f < k\} \cap B_m $$ And the bundle them up again like so $$ B'_k = \bigcup_{\ell,m \leq k} B_m^\ell $$ Notice that $B_k'$ still exhibits $\mu$ as a $\sigma$-finite measure but now $f$ is bounded on $B_k'$.
Now set $M_k = \sup_{B_k'} f$ and finish up by passing $$ \int_{B_k'} \phi_n \mathrm d\mu < M_k \mu(B_k') $$ through the monotone convergence theorem, so you can bring the limit for $n \to \infty$ (notice: $n$, not $k$) inside the integral.