Countable basis in a metric space is equivalent to being a separable metric space

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Say we have a metric space space $(\Omega, d)$. Is the following true:

$(\Omega, d)$ has a countable basis in a metric space $\iff$ $(\Omega, d)$ separable metric space

To me the equivalence seems intuitive (in terms of open balls) but I have not seen it actually stated anywhere