Show that $\lim_{n \to \infty} x_n = a \implies \lim_{n \to \infty} |x_n| = |a|$

Question

I know this is a basic question, but I could not find a question on the site about exactly this question, even by using https://approach0.xyz/search/ Maybe I searched wrongly ?

Show that $\lim_{n \to \infty} x_n = a \;\implies\; \lim_{n \to \infty} |x_n| = |a|$.

Let $(x_n)_{n \in \mathbb{N}}$ be a real sequence and $a \in \mathbb{R}$.

Show that $$ \lim_{n \to \infty} x_n = a \;\implies\; \lim_{n \to \infty} |x_n| = |a| $$

$\lim_{n \to \infty} x_n = a$ means that for each $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $\forall n > N$ there is $|x_n - a| < \epsilon$

Using the triangle inequality: $$ \epsilon > |x_n - a| \geq \bigl| |x_n| - |a| \bigr| $$

But $\epsilon > \bigl| |x_n| - |a| \bigr|$ is the definition of $\lim_{n \to \infty} |x_n| = |a|$

This completes the proof.

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Note that the reverse claim $$ \lim_{n \to \infty} x_n = a \;\Longleftarrow\; \lim_{n \to \infty} |x_n| = |a| $$

does not hold in general. For example if $x_n = (-1)^n$ then $$\lim_{n \to \infty} |x_n| = 1$$ but $$\lim_{n \to \infty} x_n \neq 1$$

The exception being for $a = 0$, because in this case \begin{align} \bigl| |x_n| - 0 \bigr| < \epsilon \; &\Longleftrightarrow \; |x_n - 0| < \epsilon \\[4pt] \lim_{n \to \infty} |x_n| = |0| \; &\Longleftrightarrow \; \lim_{n \to \infty} x_n = 0 \end{align}

Is my attempt correct?