Proof. (1) $\Rightarrow$ (2): Let $\seqi{U}\subset 2^{X}$ be an open cover of $X$. Let $\seq{V_n}\subset 2^{X}$ be a countable base for the topology on $X$, and

For any $x \in X$, there exists $i \in I$ such that $x \in U_{i}$, and $n \in \natp$ such that $x \in V_{n} \in U_{i}$. Therefore $\bigcup_{k \in K}V_{k} = X$.

Let $J \subset I$ be countable such that for each $k \in K$, there exists $j \in J$ with $V_{n} \subset U_{j}$, then $X = \bigcup_{k \in K}V_{k} = \bigcup_{j \in J}V_{j}$.

(2) $\Rightarrow$ (3): Let $n \in \nat$, then $\bracs{B(x, 1/n)|x \in X}$ is an open cover of $X$, so there exists $\seq{x_{n, k}}\subset X$ such that $X = \bigcup_{k \in \natp}B(x_{k}, 1/n)$. Let $D = \bracs{x_{n, k}| n, k \in \natp}$, then for any $x \in X$ and $n \in \natp$, there exists $k \in \natp$ such that $x \in B(x_{n, k}, 1/n)$, so $x_{n, k}\in B(x, 1/n)$. Therefore $D$ is dense in $X$, and $X$ is separable.

(3) $\Rightarrow$ (1): Let $\seq{x_n}\subset X$ be a countable dense subset. Let $x \in X$ and $k \in \natp$, then there exists $x_{n} \in \natp$ such that $d(x, x_{n}) < 1/(2k)$. In which case, $x \in B(x_{n}, 1/(2k)) \subset B(x_{n}, 1/k)$. Therefore $\bracs{B(x_n, 1/k)|n, k \in \natp}$ forms a countable basis for $X$.$\square$