This question already has an answer here:
Assume $|a+b|\leq |a|+|b|$. Then $|a-b|=|a+(-b)|\leq |a|+|(-b)|=|a|+|b|$.
$$|a-b|=|a+(-b)|\leq |a|+|-b|=|a|+|b|$$
By the triangle inequality $|a+b|\le |a|+|b|$, so also,
$$|a-b|=|a+(-b)|\le |a|+|-b|=|a|+|b|$$
If you want to prove the triangle inequality, consider proving
$$|a+b|\leq |a|+|b|$$
when $\{a\leq 0, b\geq 0$}, $\{a,b \geq 0\}$, and then reason why that would also cover the cases $\{a\geq 0, b\leq 0\}$ and $\{a,b\leq 0\}$.