Here's my proof, which I am not sure is correct :
Assume f is continuous at a
$=> \lim \limits_{x \to a} f(x) = f(a)$
$=> \lim \limits_{x \to a} f(x)$ exists
$=> \lim \limits_{x \to a} |f(x)| = | \lim \limits_{x \to a} f(x) | = | f(a) |$
Hence, |f| is continuous at a
Is there something wrong with it ?
The line $\lim | f(x) | = | \lim f(x) |$ is using the assumption of continuity of $|\cdot|$, so your argument does not work. The defining property of continuity is the statement that limits can be passed inside.
Instead, you can use the version of the triangle inequality: $$| |x| - |y| | \le |x-y|.$$
In particular, using the $\epsilon-\delta$ definition of continuity:
Suppose $\epsilon > 0$ and let $\delta > 0$ be such that for any $|x-a| < \delta$ we have $|f(x) - f(a)| < \epsilon$.
Then this means that $||f(x)| - |f(a)|| < |f(x) - f(a)| < \epsilon$ for all $|x-a| < \delta$. Hence $|f(x)|$ is continuous.
Since you might find it helpful here is a quick proof of $||x|-|y|| < |x-y|$:
$$|x| = |x-y+y| < |x-y| + |y|$$
This means $|x|-|y| < |x-y|$.
$$|y| = |y-x+x| < |y-x| + |x| = |x-y| + |x|$$
In turn this tells us that $|y|-|x| < |x-y|$ or $-|x-y| < |x|-|y|$.
Hence $$-|x-y| < |x|-|y| < |x-y|$$ and $| |x| - |y| | < |x-y|$.