Let $A \subseteq \mathbb{R}$ and $|A|=\omega$, prove $|\mathbb{R}\setminus A| = 2^\omega$.

Question

Let $A \subseteq \mathbb{R}$ and $|A|=\omega$, prove $|\mathbb{R}\setminus A| = 2^\omega$.

Hint: $\mathbb{R}\sim \{0,1\}^\mathbb{N}$

I'm somewhat stuck trying to prove this, could someone give some hints (no full answers please)

I know $|A| = \omega < |\mathcal{P}(A)|=2^{\omega}$ (Cantor).

I guess I should look for some bijective function between $\mathbb{R}\setminus A$ and $\mathbb{R}$ or $\mathcal{P}(A)$. I also don't see how I can use the hint... Should I look for a bijection between $\mathbb{R}\setminus A$ and $\{0,1\}^\mathbb{N}$?

I was given another hint to use a diagonalization argument, but isn't that just usefull to prove strict equalities like $|A| < |\mathcal{P}(A)|$?