Converse of the Urysohn metrization theorem

Urysohn metrization theorem says that every regular and second countable topological space is metrizable. My question, is the converse of this theorem ture ? If not, what are the counter examples?

Any reply kindly appreciated. Thanks.

As an example consider the set of all real numbers $\mathbb{R}$ with the discrete topology $\tau_{d}$. Clearly $(\mathbb{R},\tau_{d})$ is metrizable and the discrete metric $\rho_{d}$ is the metrization of $\tau_{d}$. Also $\mathbb{R}$ is regular with respect to $\tau_{d}$ since every closed subset of $\mathbb{R}$ is open with respect to $\tau_{d}$.
Notice that if a collection $\mathcal{B}$ of subsets of $\mathbb{R}$ is a base for $(\mathbb{R},\tau_{d})$ then $\mathcal{B}$ contains $\{\{x\}:x\in \mathbb{R}\}$. But $\{\{x\}:x\in \mathbb{R}\}$ is uncountable. Hence $\mathcal{B}$ is also uncountable. Therefore $(\mathbb{R},\tau_{d})$ has no countable base even though $(\mathbb{R},\tau_{d})$ is metrizable. Therefore the converse of the Urysohn metrization theorem does not hold.