The statement is false for sequences $\langle a_n:n\in\Bbb N\rangle$ such that $\liminf_{n\in\Bbb N}a_n=0$: $0^{-1}$ isn’t defined. It should say that
- $\limsup_{n\in\Bbb N}a_n^{-1}=(\liminf_{n\in\Bbb N}a_n)^{-1}$ when $\liminf_{n\in\Bbb N}a_n\ne 0$, and
- $\limsup_{n\in\Bbb N}a_n^{-1}=\infty$ when $\liminf_{n\in\Bbb N}a_n=0$.
Thus, boundedness of the sequence isn’t really a consideration. I’ll deal instead with the case in which $\liminf_{n\in\Bbb N}a_n=0$.
In this case we know that for each $\epsilon>0$ there is an $n_\epsilon>0$ such that $\inf\{a_k:k\ge n\}<\epsilon$ whenever $n\ge n_\epsilon$. Thus, for each $n\ge n_\epsilon$ there is a $k\ge n$ such that $a_k<\epsilon$.
We want to show that $\limsup_{n\in\Bbb N}a_n^{-1}=\infty$, i.e., that for each positive $x$ there is an $m_x\in\Bbb N$ such that $\sup\{a_k^{-1}:k\ge n\}>x$ whenever $n\ge m_x$. Let $\epsilon=\frac1x$, and let $m_x=n_\epsilon$. Then for each $n\ge m_x$ there is a $k\ge n$ such that $a_k<\epsilon$ and hence such that $a_k^{-1}>\frac1{\epsilon}=x$. This implies that $\sup\{a_k^{-1}:k\ge n\}>x$ whenever $n\ge m_x$, which is exactly what we needed.
You should take another look at what you did in the bounded case: since you didn’t correctly identify the actual trouble spot, you probably don’t have a completely correct argument.