[proofplan] We prove the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. From a countable dense set, we build a countable base of rational-radius balls centred at dense points. Second countability gives the Lindelof property by selecting one basis element inside each cover element. The Lindelof property gives separability by extracting a countable subcover from balls of radius $1/n$ for each $n$. [/proofplan] [step:$(1) \Rightarrow (2)$: Build a countable base from a countable dense subset] [claim:Separability Implies Second Countability] If $(X, d)$ has a countable dense subset $D$, then $\mathcal{B} = \{B(q, r) : q \in D,\, r \in \mathbb{Q},\, r > 0\}$ is a countable base. [/claim] [proof] $\mathcal{B}$ is countable. For any open $U$ and $x \in U$, choose $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq U$, then $q \in D$ with $d(x, q) < \varepsilon/2$, then $r \in \mathbb{Q}$ with $d(x, q) < r < \varepsilon/2$. Then $x \in B(q, r) \subseteq B(x, \varepsilon) \subseteq U$. [/proof] [/step] [step:$(2) \Rightarrow (3)$: Second countability implies the Lindelof property] [claim:Second Countability Implies Lindelof] If $X$ has a countable base $\mathcal{B} = \{B_n\}$, every open cover has a countable subcover. [/claim] [proof] For each $x$ in an open cover $\{U_\alpha\}$, choose $U_{\alpha(x)}$ and $B_{n(x)} \subseteq U_{\alpha(x)}$ with $x \in B_{n(x)}$. The set $S = \{n(x) : x \in X\} \subseteq \mathbb{N}$ is countable; for each $n \in S$ pick one $\alpha_n$. Then $\{U_{\alpha_n}\}_{n \in S}$ is a countable subcover. [/proof] [/step] [step:$(3) \Rightarrow (1)$: The Lindelof property implies separability] [claim:Lindelof Implies Separability] If every open cover has a countable subcover, then $X$ has a countable dense subset. [/claim] [proof] For each $n \in \mathbb{N}$, extract a countable subcover $\{B(q, 1/n) : q \in D_n\}$ from the cover $\{B(x, 1/n) : x \in X\}$. Set $D = \bigcup_n D_n$, which is countable. For any $x \in X$ and $\varepsilon > 0$, choose $n$ with $1/n < \varepsilon$; some $q \in D_n$ has $d(x, q) < 1/n < \varepsilon$, so $D$ is dense. [/proof] [/step]