Is the proof $|d(x, A) - d(y, A) |\le d(x, y) $ in $(X, d) $ and $\emptyset \neq A \subset X$ and $x, y\in X$ logically perfect?

Question

$(X, d) $ be a metric space and $\emptyset \neq A \subset X$ and $x, y\in X$.

To show: $|d(x, A) - d(y, A)| \le d(x, y)$

My attempt: \begin{align} &d(x, A) = \inf \{ d(x, a) : a\in A \}\\ &d(x, A) \le d(x, a) \le d(x, y) +d(y, a) \forall a\in A. \end{align}

Hence, \begin{align} d(x, A)&\le\inf\{d(x, y)+d(y, a):a\in A\}=\\ &=d(x, y) +\inf \{d(y, a):a\in A\}=\\ &=d(x, y) +d(y, A).\end{align}

It follows $$d(x, A) -d(y, A) \le d(x, y),$$ and analogously $d(y, A) \le d(y, x) +d(x, A) $, that is $$ d(y, A)-d(x, A) \le d(x, y) .$$

Then combining above two inequalities, we get

$|d(x, A) - d(y, A) |\le d(x, y) $

Is my proof correct?

Is there any logical gaps?

Please add your thoughts. Thanks.