A topological space $\mathbf{X}$ is functionally Hausdorff, if for any two distinct $x, y \in \mathbf{X}$ there exists a continuous function $f_{xy} : \mathbf{X} \to [0,1]$ with $f(x) = 0$ and $f(y) = 1$.

A space $\mathbf{X} = (X,\tau)$ is submetrizable, if there exists a topology $\tau' \subseteq \tau$ such that $(X,\tau')$ is metrizable. Equivalently, if there is a continuous injection $\iota : \mathbf{X} \to \mathbf{X}'$ to some metric space $\mathbf{X}'$.

It is rather easy to see that any submetrizable space is functionally Hausdorff. I am wondering whether restricted to second-countable spaces, the converse might hold, too.

**Failed solution attempts:**
My naive attempt to prove this was to pick a dense sequence $(a_n)_{n \in \mathbb{N}}$, and to consider the continuous map $F : \mathbf{X} \to [0,1]^\omega$ where $F(x)(\langle n,m\rangle) = f_{a_na_m}(x)$ for some tupling functions for unequal pairs. This map can fail to be injective, though: Take an uncountable space with a dense sequence of isolated points, pick the $f_{xy}$ suitably, and $$F[\mathbf{X}] = \{x \in [0,1]^\omega \mid \exists n \ \forall i \neq n \ x_n = 1 \wedge x_i = 0\} \cup \{0^\omega\}$$ is countable.

The initial topology induced by all $f_{xy}$ should be regular, and is nested between two countably-based topologies, but I do not see why it should be countably-based itself.

Searching on $\pi$-base for secound countable, functionally Hausdorff (aka Urysohn) but not metrizable spaces yields the following:

https://topology.jdabbs.com/spaces?q=Second%20Countable%20%2B%20Urysohn%20%2B%20~metrizable

Of these examples most are just defined by adding open sets to a metrizable topology. The other two (irregular lattice topology and Roy's Lattice Subspace) are countable, hence the argument above with a total enumeration shows their submetrizability.