Analysis Proof- different conditions.

A continuous function on $[a,b]$ is also uniformly continuous on $[a,b]$.

The following tries to illustrate what happens when the interval is not closed:

Show: $f(x) = \frac{1}{x}$ is not uniformly continuous in the half open interval ($0, 1$]

Proof:

Take $\epsilon = 1$. We show that there is no $\delta>0$ such that $\forall p$ and $\forall x$ such that $|p − x| < \delta$
$|f(p) − f(x)| < \epsilon$
Take sequences $x_n =\frac{1}{n}$ and $y_n =\frac{1}{n+1}$ . Then $|f(x_n) − f(y_n)| = 1$, but $|x_n − y_n| \rightarrow 0$.
So for any $\delta > 0$, there exists $n$ such that $|x_n − y_n| < \delta$ but $|f(x_n) − f(y_n)| \not< 1$. So $f$
is not uniformly continuous.

I looked at this proof, followed the definition/theorem chase, was pleased I understood, till I realised: wait... where have we used the fact this is the half open interval ($0,1$]? If I didnt know any better I would have believed this if it was a claimed disproof of the uniform continuity of $f$ on [$0,1$]
Could someone please explain?

Answer

$f$ is not defined at $0$, so it cannot be continuous at zero.