proving a  function satisfies a  Lipschitz condition 
Let $F$ be a closed set of $ \mathbb{R} $ whose complement has finite measure. Let $\delta(x) = d (x, F) =\inf \{ |x - z| \mid z \in F\}$.
Prove $ \delta$ continuous by proving $| \delta(x) - \delta(y) | \leq |x - y|$

I appreciate any kind of hint.
Thanks

Let $x, y\in\mathbb{R}$. For any $\epsilon>0$, by definition of $\delta(y)$, there exists $z\in F$ such that 
$$|y-z|\leq\delta(y)+\epsilon.$$ 
As Ashok said, we have the triangle inequality
$$|x-z|\le |x-y|+|y-z|.$$
Combining these two inequalities, we have
$$|x-z|\le |x-y|+\delta(y)+\epsilon.$$
By definiton of $\delta(x)$, we have $\delta(x)\leq |x-z|$ since $z\in F$. Hence, we have
$$\delta(x)\le |x-y|+\delta(y)+\epsilon.$$
Since $\epsilon>0$ is arbitrary, we have $\delta(x)\le |x-y|+\delta(y)$, or 
$$\delta(x)-\delta(y)\le |x-y|.$$
By symmetry, we also have
$$\delta(y)-\delta(x)\le |y-x|.$$
Therefore, we obtain
$$| \delta(x) - \delta(y) | \leq |x - y|.$$
