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
Consider the subset $K$ of $C_b(\Bbb R)$ consisting of functions that are either $0$ or $1$ at the integers. There is an uncountable subset $S$ of $K$ such that: $$\tag{1}\Vert x-y\Vert\ge 1,\ \ \text{whenever}\ \ x,y\in S\ \text{with}\ x\ne y.$$
Now, given a countable subset $B$ of $C_b(\Bbb R)$ it follows from $(1)$ that there is an $s\in S$ that is distance at least $1/2$ from every element of $B$. Then $B$ is not dense in $C_b(\Bbb R)$.