Proof for distances to a set

Question

With a metric space $(X,d)$, prove that $|d_E(x)-d_E(y)|\leq d(x,y)+d(y,z)$.

In this context, $x \in X$, $d_E(x)=\inf\left\{d(x,z) : z \in E\right\}$, E is a subset of X.

I've already proved the statement without absolute values; I tried using contradiction and the definition of absolute value, but it didn't really lead anywhere.

Answer 1

$$ d_E(x)\leq d(x,z)\leq d(x,y)+d(y,z) \quad \forall z\in E $$ Thus,$d_E(x)-d(x,y)\leq d_E(y)\Rightarrow d_E(x)-d_E(y)\leq d(x,y)\leq d(x,y)+d(y,z)$. The last inequality holds since $d(.,.)\geq0$.

Similarly, $$ d_E(y)\leq d(y,z)\leq d(y,x)+ d(x,z) \quad \forall z\in E $$

So, $d_E(y)-d(y,x)\leq d_E(x)\Rightarrow -d(x,y)-d(y,z)\leq-d(y,x)\leq d_E(x)-d_E(y)$.Then argument follows at once from these two.

Answer 2

$$\vert d_E(x) - d_E(y) \vert \leq \max\{d_E(x),d_E(y)\} \leq \max\{d(x,z),d(y,z)\} \leq d(x,y) + d(y,z).$$

Here the last inequality uses the triangle inequality