Show that if $(x_n)$ is unbounded then $\exists$ subsequence $x_{n_{k}}$ such that $lim (1/x_{n_{k}})=0$

Question

A proof to the statement in title is already here but I am trying to prove it by contradiction. The following is my proof:
Statement: If $(x_n)$ is unbounded then $\exists$ subsequence $x_{n_{k}}$ such that $lim (1/x_{n_{k}})=0$
Suppose that for all subsequences of $x_n$, the implication is not true.
$\therefore$ for any subsequence ($x_{n_{k}}$), ($\exists \epsilon\gt0$)($\forall K\in N$)($\exists n_k\ge K$): $|1/x_{n_k}|\gt \epsilon \Rightarrow |x_{n_k}|\lt \epsilon$ (A)
(A) is true for all subsequences $\Rightarrow$ all subsequences of $x_n$ are bounded
Now my claim is: All subsequences of $x_n$ are bounded $\implies$ $x_n$ is bounded
Proof: Let M= sup(set of bounds of all subsequences), if $x_n$ is unbounded then $(\exists N)(\forall n\ge N): |x_n|\gt M$. Now subsequence $(x_{m_n}): |x_{m_n}|\lt M$ which is a contradiction since $m_n\ge n$
$\therefore$ (A) is true for all subsequences $\Rightarrow$ all subsequences of $x_n$ are bounded $\implies$ $x_n$ is bounded, which is a contradiction.
Is my proof correct? If wrong,please advise on how to prove the statement using contradiction.Thanks in advance.