Show for $x,y \in \mathbb{X} : |d_A (x) - d_A (y)| \leq d(x,y)$ For $A \subset \mathbb{X}$ non empty and $x \in \mathbb{X}$ define the distance of $x$ to $A$ by $$d_A (x) =  \inf \limits _{a \in A} d(x,a)$$
I am trying to show for $$x,y \in \mathbb{X} : |d_A (x) - d_A (y)| \leq d(x,y)$$
This is the proof I have.
I start with the triangle inequality: $$d(x,a) \leq d(x,y) + d(y,a)$$
I note $$d_A (x) = \inf \limits _{a \in A} d(x,a)$$ and $$d_A (y) = \inf \limits _{a \in A} d(y,a)$$
So the triangle inequality becomes $$d_A (x) \leq d(x,y) + d_A (y)$$
At this point it is obvious you rearrange to get the desired solution but that was marked as wrong and instead you change the role of $x$ and $y$. 
Why is rearranging wrong? 
And why do you have to change the roles of $x$ and $y$?

In order to prove that $$\vert d_A (x) - d_A (y) \vert \leq d(x,y),$$ you have to prove the following two inequalities:
$$\begin{cases}
-d(x,y) \le d_A (x) - d_A (y) \\
d_A (x) - d_A (y) \le d(x,y)
\end{cases}.$$ In the question of your post, you proved the second one.
Rearranging only won't be enough to obtain the first one. However, as $x,y$ play symmetric roles, you can permute them to get the first inequality.
