How to prove that $f(x) =1/x$ is unbounded in $(0,1)$?

By definition, $f$ is bounded if and only if there exists $M\in\mathbb{N}$ such that for all $x\in (0,1)$, we have $|f(x)| < M$.

Assume that such an $M$ exists, now take $x=\frac{1}{M+1}$. Then $f(x)=M+1$ and $|f(x)| > M$. Contradiction.

It's lower bound is obviously $1$, it is a constantly decreasing function in the range and $f(1)=1$ is defined.

For the upper bound, assume that bound is $U=\frac{1}{x_0}$

Then evaluate $f(\frac{x_0}{2})$, we know $\frac{x_0}{2}<x_0$. Then apply the knowledge that for $a,b>0$ we have $$a>b\iff\frac ab> 1$$ What does this tell you?