Proof of $|x|$ being continuous

Question

I have attempted to prove that $f(x)=|x$| is continuous at all points $a\in \mathbb{R}$, is this correct?

Scratch work:
$|f(x)-f(a)|=||x|-|a||$
By the reverse triangle ineqaulity $||x|-|a|| \le |x-a|$ so if we have that $|x-a|<\epsilon$ this will prove that $|x|$ is continuous.

Proof: $\epsilon > 0$ is given, we choose $\delta =\epsilon$. Then $x \in \mathbb{R}$ satisfying $|x-a| <\delta$ is given.

Now $|f(x) -f(a)|$ = $||x|-|a|| \le |x-a| < \delta = \epsilon$.

Answer 1

Your proof is fine. Excellent.

One more proof : $f : \mathbb R \rightarrow \mathbb R$ is continuous iff inverse image of $(a,b)$ is open for each $a,b \in \mathbb R $ and $a<b$. we are cosidering usual distance metric on $\mathbb R$.

$f^{-1}(a,b)=(a,b) \cup(-b,-a)$ which is open in $\mathbb R$.

Hence $f$ is continuous.