Is this discontinuity proof finished?

Solution 1:

I assume that you are working from the definition of continuity, or rather from its negation, and so you must prove:

There exists $\epsilon > 0$ such that for all $\delta > 0$ there exists $x$ such that $|x-x_0| < \delta$ and $|f(x)-f(x_0)| \ge \epsilon$.

You've made a good start by choosing $\epsilon = \frac{1}{2}$. But no, you're not done.

What you must still do is: let $\delta > 0$, and then find an appropriate value of $x$ (depending on $\delta$) and use it to prove that $|x-x_0| < \delta$ and that $|f(x)-f(x_0)| \ge \epsilon$.

With the formula you worked out for $|f(x)-f(x_0)|$, I suspect you can now easily find $x$.