Let $A$ be the collection of all bounded analytic functions on unit disc $\mathbb{D} = \{z \in \mathbb{C} : |z| <1\}$ with sup norm.
I have done showing $A$ is Banach : If $(f_n)$ is Cauchy sequence, then for any $z$, $(f_n(z))$ is Cauchy, so there exist a pointwise limit $f(z)$. Now, it is easy to show that $f$ is bounded . To show that $f$ is analytic we use, uniform convergence and Morera's theorem and then we prove $f_n \to f$. This is all about showing $A$ is Banach space.
Now, I am interested in showing whether this space is separable or not.
I did the following argument:
Let $E$ be a countable dense subset of $\mathbb{D}$. Any $f \in A$ is also continuous, and thus all sequences in $E^{\mathbb{N}}$ uniquely determine $f$, so there is a bijection from $E^{\mathbb{N}}$ to $A$. So cardinality of $A$ is cardinality of $E^\mathbb{N}$ is same as cardinality of $\mathbb{R}$. So, $|A| = |\mathbb{R}|$. Now, this does not prove that $A$ is separable, but we also know that separable spaces have cardinality of at most $|\mathbb{R}|$ .So this is just a possibility.
The other ways I thought to apply Stone Weierstrass theorem, but I am not sure if that is applicable to complex functions. How do I prove that $A$ is separable or not?