Your work seems right (although you could get the easier contradiction in the second case by adding $d(x,y)$), but there is an easier proof: Let $x,y,z$ be points in a metric space. Then
$$d(x,y)\leq d(x,z)+d(z,y).$$
Then,
$$d(x,y)-d(y,z)\leq d(x,z).$$
On the other hand,
$$d(y,z)\leq d(x,z)+d(x,y).$$
Then,
$$d(y,z)-d(x,y)\leq d(x,z).$$
Therefore,
$$|d(y,z)-d(x,y)|\leq d(x,z).$$
A nice proof can be found in chapter 2 of The Cauchy-Schwarz Master Class by J.M. Steele. It is stated in form of some problems starting with:
Problem 2.1: Show that for every sequence of nonnegative real numbers $a_1,a_2,\ldots,a_n$ one has
\begin{align*}
(a_1a_2\cdots a_n)^{1/n}\leq \frac{a_1+a_2+\cdots a_n}{n}\tag{1}
\end{align*}
The proof of this and the following problems is done in some steps.
Step 1: $n=2^k, k\geq 1$
From the Cauchy-Schwarz inequality
\begin{align*}
\sqrt{a_1a_2}\leq \frac{a_1+a_2}{2}\tag{2}
\end{align*}
which is (1) with $n=2$ we can go on one step to $n=4$ by applying (2) twice
\begin{align*}
(a_1a_2a_3a_4)^{\frac{1}{4}}\leq \frac{(a_1a_2)^{\frac{1}{2}}+(a_3a_4)^{\frac{1}{2}}}{2}
\leq \frac{a_1+a_2+a_3+a_4}{4}
\end{align*}
This new bound can be used to prove (1) for $n=8$
\begin{align*}
(a_1a_2\cdots a_8)^{\frac{1}{8}}\leq \frac{(a_1a_2a_3a_4)^{\frac{1}{4}}+(a_5a_6a_7a_8)^{\frac{1}{4}}}{2}
\leq \frac{a_1+a_2+\cdots+a_8}{8}
\end{align*}
Iteratively going on this way we can show AM-GM for $n=2^k,k\geq 1$.
\begin{align*}
\color{blue}{(a_1a_2\cdots a_{2^k})^{\frac{1}{2^k}}\leq\frac{a_1+a_2+\cdots+a_{2^k}}{2^k}}
\end{align*}
Step 2: $2^{k-1}<n<2^k$
The next step is to fill the gaps between $2^{k-1}$ and $2^k$. This is done by taking a value $n<2^k$, considering $\alpha_i=a_i$ for $1\leq i\leq n$ and setting
\begin{align*}
\alpha_i=\frac{a_1+a_2+\cdots+a_n}{n}\equiv A\qquad\qquad n<i\leq 2^k
\end{align*}
The sequence $\{a_i:1\leq i\leq n\}$ is transformed to a sequence $\{\alpha_i:1\leq i\leq 2^k\}$ padded with $2^k-n$ copies of the average $A$ and it can be shown
\begin{align*}
\color{blue}{(a_1a_2\cdots a_n)^{1/n}\leq \frac{a_1+a_2+\cdots+a_n}{n}}
\end{align*}
Step 3: rational numbers
Based upon step (2) the AM-GM is shown for non-negative rational numbers $p_1,p_2,\ldots,p_n$ that sum to one.
\begin{align*}
\color{blue}{a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq p_1a_1+p_2a_2+\cdots +p_na_n}
\end{align*}
Step 4: real numbers
Based upon step (3) the AM-GM is shown for non-negative real numbers $p_1,p_2,\ldots,p_n$ that sum to one by taking limits.
\begin{align*}
\color{blue}{a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}\leq p_1a_1+p_2a_2+\cdots +p_na_n}
\end{align*}
Note: All these steps are some kind of bootstrapping based upon the step before and starting with the Cauchy-Schwarz inequality.
Best Answer
Don't work too hard: $|x| - |y| \geq -|x - y|$ is true because $|x| + |y - x| \geq |y|$,
and $|x| - |y| \leq |x - y|$ because $|y| + |x - y| \geq |x|$ . These are true because of triangle inequality, hence the answer.