Where $H:\mathbb{R} \to \mathbb{R}$ is defined by $H(x)=\begin{cases}1&\quad \text{if }x\text{$\geq$0}\\0&\quad\text{if } x \text { otherwise}\end{cases}$. Prove that for all $a\neq0$ the function $H$ is continuous at $a$. Also prove that $H$ is not continuous at 0.
I tried where $\delta = \epsilon$ but that doesn't work for $a=1/3$ $\epsilon=2/3$ $x=-1/4$. With a correct $\delta$ proving it is continuous at $a$ I should be able to do but not sure how to prove something is not continuous at a point, have never done that before.
For $ a>0$ take $\delta =a$ and verify that $|x-a| <\delta$ implies $x>0$ so $H(x)=H(a)=1$ and $|H(x)-H(a)| <\epsilon$. For $ a<0$ take $\delta =-a$ and verify that $|x-a| <\delta$ implies $x<0$ so $H(x)=H(a)=0$ and $|H(x)-H(a)| <\epsilon$.
Suppose $H$ is continuous at $0$. There exists $\delta >0$ such that $|H(x)-H(0)| <1$ if $|x-0| <\delta$. Take $x =-\delta /2$ to see that $|0-1| <1$ which is a contradiction .