To be a metric, $d$ must satisfy the following three conditions:
- $d(x,y) \geq 0$ for all $x$, $y \in X$.
- $d(x,y) = 0$ if and only if $x=y$.
- $d(x,y) = d(y,x)$ for all $x$, $y\in X$.
- $d(x,z) \leq d(x,y) + d(y,z)$ for all $x$, $y$, $z \in X$ (the triangle inequality)
Take these conditions in turn. Beforehand, note that for any $x$, $y \in X$ we have $d(x,y) \geq d_1(x,y)$ and $d(x,y) \geq d_2(x,y)$.
$d(x,y) \geq d_1(x,y) \geq 0$ for any $x, y \in X$.
If $x \neq y$ then $d_1(x,y) >0$ so $d(x,y) \geq d_1(x,y) >0$. If $x=y$ then $d_1(x,y) = d_2(x,y) =0$ so therefore their maximum $d(x,y)$ is also zero.
$d(x,y) = \max\{d_1(x,y),d_2(x,y)\} = \max\{d_1(y,x),d_2(y,x)\} = d(y,x)$.
Given $x, y, z \in X$, $d(x,z)$ is either equal to $d_1(x,z)$ or $d_2(x,z)$. Suppose first that $d(x,z) = d_1(x,z)$. By the triangle inequality for $d_1$ we have $d(x,z) = d_1(x,z) \leq d_1(x,y) + d_1(y,z) \leq d(x,z) + d(y,z)$. Similarly the triangle inequality holds for this triple $x, y, z$ if $d(x,y) = d_2(x,y)$.
Therefore $d$ is a metric on $X$.