I shall answer question 1 here by proving by contrapositive that every metric space with countable extent is Lindelof. Assume that $X$ is a metric space that is not Lindelof. Then $X$ is not separable (for metric spaces, the properties of second countability, Lindelof, and separability are all equivalent as commented above. See Dugundji p. 187). We shall now construct a sequence $(x_{\alpha})_{\alpha<\omega_{1}}$ by transfinite induction. For each $\alpha<\omega_{1}$, let $U_{\alpha}=X\setminus\overline{\{x_{\beta}|\beta<\alpha\}}$. Clearly $U_{\alpha}$ is a non-empty open set for all $\alpha$ since $X$ is not separable. Let $\epsilon_{\alpha}=\sup\{\epsilon|B_{\epsilon}(x)\subseteq U_{\alpha}\,\textrm{for some}\,x\in X\}$. Let $x_{\alpha}$ be a point where $B_{\epsilon_{\alpha}/2}(x_{\alpha})\subseteq U_{\alpha}$ for all $\alpha$. We observe that $\epsilon_{\alpha}$ is a decreasing sequence of positive real numbers of length $\omega_{1}$. Therefore, the sequence $\epsilon_{\alpha}$ is eventually some constant $\epsilon$. \epsilon>0$. Therefore$B_{\epsilon/2}(x_{\alpha})\subseteq U_{\alpha}=X\setminus\overline{\{x_{\beta}|\beta<\alpha\}}$, so if$\beta<\alpha$, then$d(x_{\beta},x_{\alpha})\geq\epsilon/2$. Therefore the set$\{x_{\alpha}|\alpha<\omega_{1}\}$is a closed discrete setset(in fact uniformly discrete), so$X$does not have countable extent. 1 I shall answer question 1 here by proving by contrapositive that every space with countable extent is Lindelof. Assume that$X$is a metric space that is not Lindelof. Then$X$is not separable (for metric spaces, the properties of second countability, Lindelof, and separability are all equivalent as commented above. See Dugundji p. 187). We shall now construct a sequence $(x_{\alpha})_{\alpha<\omega_{1}}$ by transfinite induction. For each$\alpha<\omega_{1}$, let$U_{\alpha}=X\setminus\overline{\{x_{\beta}|\beta<\alpha\}}$. Clearly$U_{\alpha}$is a non-empty open set for all$\alpha$since$X$is not separable. Let $\epsilon_{\alpha}=\sup\{\epsilon|B_{\epsilon}(x)\subseteq U_{\alpha}\,\textrm{for some}\,x\in X\}$. Let$x_{\alpha}$be a point where$B_{\epsilon_{\alpha}/2}(x_{\alpha})\subseteq U_{\alpha}$for all$\alpha$. We observe that$\epsilon_{\alpha}$is a decreasing sequence of positive real numbers of length$\omega_{1}$. Therefore, the sequence$\epsilon_{\alpha}$is eventually some constant$\epsilon$. Therefore$B_{\epsilon/2}(x_{\alpha})\subseteq U_{\alpha}=X\setminus\overline{\{x_{\beta}|\beta<\alpha\}}$, so if$\beta<\alpha$, then$d(x_{\beta},x_{\alpha})\geq\epsilon/2$. Therefore the set$\{x_{\alpha}|\alpha<\omega_{1}\}$is a closed discrete set, so$X\$ does not have countable extent.