Show that f is not continuous by epsilon-delta

Question

Given $\,f:\mathbb{R} \rightarrow \mathbb{R}$ by $f(x)$ \begin{cases} x+1 &\mbox{if} \,\,x \le 3 \\2 &\mbox{if} \,\,x > 3 \end{cases}

Use the $\epsilon-\delta$ definition of continuity to show that $f$ is not continuous at $x=3$.

What I have done so far: To show $f$ is not continuous, it is the same as $\exists\epsilon > 0, \forall\delta > 0, \exists x \,\,\mbox{such that} \,\,|x-3| < \delta \,\,\mbox{and}\,\, |f(x) - f(3)| > \epsilon$

Chose $\epsilon = 0.5$

Check the case $x>3$, $|f(x) - f(3)| = |2-4| = 2 > \epsilon$. QED

I am not sure if this is sufficient for the proof or not. Could someone please help with this? Thanks

Answer 1

Seems fine to me.

If you want, you can pick $x=3+\frac{\delta}{2}$ explicitly.