A Proposed Converse to A Theorem on Products of Limits

Yes.

Suppose that $(x_n)$ is not bounded. We construct a sequence $(y_n)$ that converges to $0$ but for which $(x_ny_n)$ does not converge to $0$.

Since $(x_n)$ is not bounded, given any $N\gt 0$ and $m\gt 0$, there exists $n\gt m$ such that $|x_n|\gt N$ (there must be infinitely many terms with absolute value greater than $N$).

So, first, let $n_1$ be such that $|x_{n_1}|\gt 1$. Let $n_2\gt n_1$ be such that $|x_{n_2}|\gt 2$. Assuming $n_k$ has been defined such that $|x_{n_k}|\gt k$, let $n_{k+1}\gt n_k$ be such that $|x_{n_{k+1}}|\gt k+1$.

Now define $y_n$ to be $0$ if $n\notin \{n_k\mid k=1,2,3\ldots\}$, and $y_{n_k}=\frac{1}{x_{n_k}}$.

Note that since $|x_{n_k}|\to\infty$, then $y_n\to 0$. However, the sequence $(x_ny_n)$ consists of $0$ in all indices different from the $n_k$, and $1$s in the $n_k$ terms; this sequence either does not converge at all, let alone to $0$, or converges to $1$ (as it has a subsequence that is constant $1$).

Yes. Suppose $(x_n)_n$ is an unbounded sequence. Let $s_n = \max(1,\max_{i \in [1,n]} |x_n|)$. Then $y_n = 1/s_n$ gives a $(y_n)_n$ where $\lim_{n \rightarrow \infty}y_n = 0$, so is in the set of sequences converging to zero. For this $(y_n)_n$, $\limsup_{n \rightarrow \infty} |x_n y_n| = 1$, so $\lim_{n \rightarrow \infty} x_n y_n$, if it exists, is not $0$. By contraposition, we must have that $(x_n)_n$ is a bounded sequence.