Show that $|f |\colon A \to \mathbb R$ defined by $| f | = |f(x)|$ is continuous at $a$

Question

Let $f\colon A \to \mathbb R$ be continuous at $a\in A$. Show that $| f |\colon A \to \mathbb R$ defined by $| f |(x) = | f(x)|$ is continuous at $a$.

Given $\epsilon > 0$. And for all $\epsilon > 0$ there exists $\gamma > 0$. I know that since $f\colon A \to \mathbb R$ is continuous at $a\in A$ then it follows that $\mid x - a\mid < \gamma$ and $\mid f(x) - f(a) \mid < \epsilon$

$\mid f(x) - f(a) \mid = \mid 2f(x) - f(a) - f(x) \mid = \mid 2f(x)\mid - \mid f(a) + f(x)\mid$

Ok guys, after bashing my head I realised that I over complicated the whole thing and noticed what to do. You can ignore my reasoning above.

Since $f$ is continuous at $a$ all $\epsilon > 0$ and there exists $\gamma > 0$ such that $ x \in A$

This is what I really missed:

$\mid \mid f(x) \mid - \mid f(a) \mid \mid \le \mid f(x) - f(a) \mid $

so

$\mid \mid f(x) \mid - \mid f(a) \mid \mid \le \mid f(x) - f(a) \mid \lt \epsilon $

hence

$\mid \mid f(x) \mid - \mid f(a) \mid \mid \lt \epsilon $

therefore $\mid f \mid$ is continuous at $a$ also.

Took forever but the questions did help me. Always surprising to see how simple it really is. Thanks!