1/x on (0,1] is not uniformly continuous

1. Not uniformly continuous. Take $x_n=\frac1n$ and $y_n=\frac1{2n}$ in $(0,1]$. Then $|x_n-y_n|=\frac1{2n}\to0$, but
$$|f(x_n)-f(y_n)|=|n-2n|=n\ge1.$$
Thus $\varepsilon_0=1$ defeats every $\delta$.

2. On $[\frac1{10},1]$. For $x,y\ge\frac1{10}$,
$$\left|\frac1x-\frac1y\right|=\frac{|x-y|}{xy}\le100\,|x-y|,$$
so $f$ is $100$-Lipschitz, hence uniformly continuous.

The theorem. Heine's theorem states that a continuous function on a closed bounded interval $[a,b]$ is automatically uniformly continuous. The problem on $(0,1]$ is that the interval is not closed: the function misbehaves near the missing endpoint $0$.

Remark. A continuous function on $(0,1]$ is uniformly continuous if and only if it has a finite limit at $0$, that is, if and only if it extends continuously to $[0,1]$.