Let $X$ be a space, $\ f:X\times X\rightarrow\mathbb{R}^+\cup\{0\}$ be a map satisfying the first two axioms for a metric (so $f(x,y)=0$ exactly when $x=y$, and $f$ is symmetric). Now, consider the topology generated by the following sets (informally thought of as open balls), for $x\in X$ and $a\in \mathbb{R}$: $$ U(x,a)=\{y\in X: f(x,y)<a\}. $$ Assuming this topology can be proven to be Hausdorff, second-countable and regular, Urysohn's metrization theorem gives that $X$ is metrizable.

From this information, can one conclude that $f$ is a metric?