I would like to show that the function $\frac{1}{x}$ is not uniformly continuous on $(0,1)$ using two approaches.
First Approach:
We have the fact that if a function $f$ is uniformly continuous on an open interval $(a,b)$, then the function $f$ is bounded on $(a,b)$. By using its contrapositive, since $\frac{1}{x}$ is not bounded on $(0,1)$, it is not uniformly continuous.
Second Approach:
Proof by definition. Note that the definition of non-uniform continuity is:
There exists an $\epsilon_0>0$ such that for all $\delta>0$, there exists $x,y \in (0,1)$ such that $|x-y|<\delta$ but $|\frac{1}{x} - \frac{1}{y}| \geq \epsilon_0$.
Assume that $\delta < 1$. Let $x = y + \frac{\delta}{2} \in (0,1), y < \frac{\delta}{2} \in (0,1), \epsilon_0 = 1$. Then we have
$$|\frac{1}{x} - \frac{1}{y}| = |\frac{1}{y + \frac{\delta}{2}} - \frac{1}{y}| = \frac{\frac{\delta}{2}}{|y(y + \frac{\delta}{2})|} > \frac{\frac{\delta}{2}}{(\frac{\delta}{2})(\delta)} = \frac{1}{\delta} > 1 = \epsilon_0 $$
If $\delta \geq 1$, then let $\delta^{\prime} = \frac{1}{\delta}$ so that $\delta^{\prime} \leq 1$. The rest proceeds the same as above.
Question: Are my two approaches correct?
Here's a third approach: If $f:(0,1)\to \mathbb R$ was uniformly continuous it had a continuous extension $f:[0,1]\to \mathbb R$ which is a contradiction for the case $f(x)=1/x$.
(Note that Cauchy sequences are preserved under uniformly continuous maps, this enables you to build the extension).