Let $x$ and $y$ be real numbers. Then $|x+y| ≤ |x|+|y|$ .

I am trying to prove the Triangle Inequality using both of the following Theorem.

Theorem 5.14a: $-|x| \le x \le |x|$ for all $x \in \mathbb{R}$.

Theorem 5.14b Let $a$ be a real number such that $a \ge 0$. $|x| \le a$ if and only if $-a \le x \le a$ for all $x \in \mathbb{R}$.

The following is my proof.

Proof. From Theorem 5.14a,

\begin{equation}
-|x+y| \le x+y \le |x+y|
\end{equation}

Because both $|x|$ and $|y|$ are non-negative real numbers, $|x|+|y|$ is the non-negative real number $x+y$.

It follows from Theorem 5.14b that

\begin{equation}
-(x+y) \le x+y \le x+y
\end{equation}

For the inequalities $-|x+y| \le x+y \le |x+y|$ and $-(x+y) \le x+y \le x+y$ to hold true $x+y \ge 0$.

By dividing the above inequalities by $x+y$, one obtains

\begin{equation}
-1 \le 1 \le 1
\end{equation}

Thus, given that both $x$ and $y$ are real numbers $|x+y| ≤ |x|+|y|$.

Is the proof correct?

Reference

Daepp, U. and Gorkin, P., 2011. Reading, Writing, and Proving. 2nd ed. pp.55.