[proofplan] We prove the equivalence by establishing the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$. For $(1) \Rightarrow (2)$: a second-countable space is separable by [Second-Countable Implies Separable](/theorems/1067), which holds in arbitrary topological spaces without metrizability. For $(2) \Rightarrow (3)$: from a countable dense set $D$ in a metric space, the rational-radius balls centred at points of $D$ form a countable base, reducing to the case of [Second-Countable Implies Lindelof](/theorems/1104). For $(3) \Rightarrow (1)$: the Lindelof property allows us to extract, for each $n \in \mathbb{N}$, a countable set $D_n$ whose $1/n$-balls cover $X$; the union $D = \bigcup_{n=1}^\infty D_n$ is a countable dense set, and the rational-radius balls centred at $D$ form a countable base. [/proofplan]

[step:$(1) \Rightarrow (2)$: Deduce separability from second countability] Since $(X, d)$ is second-countable, it is separable by [Second-Countable Implies Separable](/theorems/1067). This implication holds for arbitrary topological spaces and does not require metrizability. [/step]

[step:$(2) \Rightarrow (3)$: Build a countable base from a countable dense set, then apply Lindelof]Let $D \subset X$ be a countable dense subset. Define the collection \begin{align*} \mathcal{B} := \{B(q, r) : q \in D,\; r \in \mathbb{Q},\; r > 0\}, \end{align*} where $B(q, r) := \{x \in X : d(x, q) < r\}$ is the open ball of radius $r$ centred at $q$. [claim:RationalBallBase] The collection $\mathcal{B}$ is a countable base for the metric topology on $X$. [/claim] [proof] **Countability.** The set $D$ is countable and $\mathbb{Q}_{>0}$ is countable, so $\mathcal{B}$ is indexed by the countable product $D \times \mathbb{Q}_{>0}$, hence countable. **Base property.** Let $G \subset X$ be open and let $x \in G$. Since $G$ is open in the metric topology, there exists $\varepsilon > 0$ with $B(x, \varepsilon) \subset G$. Since $D$ is dense in $X$, there exists $q \in D$ with $d(x, q) < \varepsilon / 2$. Choose $r \in \mathbb{Q}$ with $d(x, q) < r < \varepsilon / 2$. Then $x \in B(q, r)$ (since $d(x, q) < r$). For any $y \in B(q, r)$, the triangle inequality gives \begin{align*} d(y, x) \le d(y, q) + d(q, x) < r + r < \varepsilon, \end{align*} so $y \in B(x, \varepsilon) \subset G$. Therefore $B(q, r) \subset G$, confirming that $x \in B(q, r) \subset G$ with $B(q, r) \in \mathcal{B}$. [/proof] Since $\mathcal{B}$ is a countable base, the space $X$ is second-countable. By [Second-Countable Implies Lindelof](/theorems/1104), $X$ is Lindelof.[/step]

[guided]Assume $X$ contains a countable dense subset $D$. We exploit the metric structure to convert density into a countable base. Define \begin{align*} \mathcal{B} := \{B(q, r) : q \in D,\; r \in \mathbb{Q},\; r > 0\}. \end{align*} Why rational radii? We need countably many radii, and $\mathbb{Q}_{>0}$ is a countable dense subset of $(0, \infty)$. Using all positive reals as radii would produce an uncountable family; restricting to rationals keeps the family countable while losing nothing, because the density of $\mathbb{Q}$ in $\mathbb{R}$ lets us always find a rational radius between any two real numbers. The claim above verifies that $\mathcal{B}$ is a base. The key step in the verification uses both the density of $D$ (to find a centre $q$ near $x$) and the density of $\mathbb{Q}$ in $\mathbb{R}$ (to find a rational radius $r$ satisfying $d(x,q) < r < \varepsilon/2$). The triangle inequality then ensures $B(q,r) \subset B(x, \varepsilon) \subset G$. With a countable base in hand, the Lindelof property follows from [Second-Countable Implies Lindelof](/theorems/1104), which is a purely topological result. Note that this step --- building a countable base from a dense set --- is the only place in the entire cycle where the metric is used in an essential way. The metric provides the open balls and the triangle inequality. In a general (non-metrizable) topological space, separability does not imply second countability: the Sorgenfrey line is separable but not second-countable.[/guided]

[step:$(3) \Rightarrow (1)$: Extract a countable dense set from the Lindelof property, then build a countable base]Assume $X$ is Lindelof. For each $n \in \mathbb{N}$, the collection $\{B(x, 1/n) : x \in X\}$ is an open cover of $X$. By the Lindelof property, it has a countable subcover: there exists a countable set $D_n \subset X$ with \begin{align*} X = \bigcup_{q \in D_n} B(q, 1/n). \end{align*} Define $D := \bigcup_{n=1}^\infty D_n$. This is a countable union of countable sets, hence countable by [Countable Union of Countable Sets](/theorems/755). [claim:LindelofDenseSet] The set $D$ is dense in $X$. [/claim] [proof] Let $x \in X$ and $\varepsilon > 0$. Choose $n \in \mathbb{N}$ with $1/n < \varepsilon$. Since $\{B(q, 1/n)\}_{q \in D_n}$ covers $X$, there exists $q \in D_n$ with $x \in B(q, 1/n)$, i.e., $d(x, q) < 1/n < \varepsilon$. Since $q \in D_n \subset D$, the ball $B(x, \varepsilon)$ contains a point of $D$. [/proof] Since $D$ is a countable dense subset, the collection $\mathcal{B} := \{B(q, r) : q \in D,\; r \in \mathbb{Q},\; r > 0\}$ is a countable base for the metric topology, by the same argument used in the implication $(2) \Rightarrow (3)$. Therefore $X$ is second-countable.[/step]

[guided]Assume every open cover of $X$ has a countable subcover. We must produce a countable base for the metric topology. The strategy is to first extract a countable dense set from the Lindelof property, then convert it into a countable base using rational-radius balls (as in the previous step). **Extracting a countable dense set.** For each $n \in \mathbb{N}$, the family $\{B(x, 1/n) : x \in X\}$ is an open cover of $X$ (every point $x$ belongs to $B(x, 1/n)$). By the Lindelof property, there exists a countable subset $D_n \subset X$ such that \begin{align*} X = \bigcup_{q \in D_n} B(q, 1/n). \end{align*} Define $D := \bigcup_{n=1}^\infty D_n$. By [Countable Union of Countable Sets](/theorems/755), $D$ is countable. Why is $D$ dense? Let $x \in X$ and $\varepsilon > 0$. Choose $n \in \mathbb{N}$ with $1/n < \varepsilon$ (possible by the Archimedean property of $\mathbb{R}$). The cover $\{B(q, 1/n)\}_{q \in D_n}$ includes $x$ in some ball $B(q, 1/n)$, so $d(x, q) < 1/n < \varepsilon$ and $q \in D_n \subset D$. Hence every open ball around $x$ meets $D$, confirming $\overline{D} = X$. **Building the countable base.** With a countable dense set $D$ in hand, the same construction from the implication $(2) \Rightarrow (3)$ applies: the rational-radius balls $\mathcal{B} = \{B(q, r) : q \in D,\; r \in \mathbb{Q}_{>0}\}$ form a countable base for the metric topology. Therefore $X$ is second-countable. This completes the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$, establishing the equivalence of all three conditions.[/guided]