Let $X$ be a normed linear space. Show that a norm $\|\cdot\|_{1}$ is stronger than a norm $\|\cdot\|_{2}$ if and only if for any sequence $\{x_{n}\} \subset X$, $\|x_{n}\|_{1} \to 0$ always implies $\|x_{n}\|_{2} \to 0$.
My work:
$\Longrightarrow$ Suppose $\|\cdot\|_{1}$ is stronger than $\|\cdot\|_{2}$. This means that there exists some $M > 0$ such that $\|x\|_{2} \leqslant M\|x\|_{1}$ for all $x \in {X}$. Let $\{x_{n}\} \subset X$ be any sequence such that $\|x_{n}\|_{1} \to 0$. It follows that $M\|x_{n}\|_{1} \to 0$ which necessarily implies $\|x_{n}\|_{2} \to 0$. (Can someone verify this?)
$\Longleftarrow$ Suppose that $\|x_{n}\|_{1} \to 0$ always implies $\|x_{n}\|_{2} \to 0$. This implies that $x_{n}$ is a Cauchy sequence under both norms. Thus, for some $\epsilon_{1}, \epsilon_{2} > 0$, there exists $N_{1}, N_{2} > \mathbb{N}$ such that $\|x_{n} - x_{m}\|_{1} < \epsilon_{1}$ and $\|x_{n} - x_{m}\|_{2} < \epsilon_{2}$ for all $n,m > N_{1},N_{2}$, respectively. ....
My question is for the reverse direction, how do I connect this idea of a "stronger" norm knowing only that both norms converge to $0$.