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? 



When is a real-valued function on a metric space a metric?
Following the example of my comment, with $X = \mathbb{R}^2$ and 


$$f((x\_1, y\_1), (x\_2, y\_2)) = ((x\_1 - x\_2)^{1/2} + (y\_1 - y\_2)^{1/2})^2,$$ 


one sees that $f((0, 0), (0, 1/2)) = 1/2$ and $f((0, 1/2), (1/2, 1/2)) = 1/2$, whereas $f((0, 0), (1/2, 1/2)) = 2$, so the triangle inequality fails. 


In general, if $X$ is a vector space (over a local field such as $\mathbb{R}$) and $N: X \to [0, \infty)$ satisfies the homogeneity condition $N(\alpha x) = |\alpha|N(x)$ for scalars $\alpha$, then the function $f(x, y) = N(x - y)$ is a metric iff the unit ball $\{x: N(x) \leq 1\}$ is convex; this implies that every ball is convex. So in some sense the triangle inequality is a generalized convexity condition. 


