Let $X$ normed space with two norms $(X,|\cdot|_1),(X,|\cdot|_2)$.
Suppose that for every $x_n,x \in X$ such that $|x_n-x|_2\to 0$ then $|x_n-x|_1\to 0$ so that the convergence induced by $|\cdot|_2$ is stronger then the one induced by $|\cdot|_1$ and hence the topology induced by $|\cdot|_2$ is stronger.
Now in this framework is it possible to prove
$$|x|_1 \leq C |x|_2$$
so that also the norm $|\cdot|_2$ is stronger then $|\cdot|_1$?
To me it seems not the case.
Suppose by contradiction that $\sup_{x\ne 0}\|x\|_1/\|x\|_2=\infty.$ Then for each $n\in \Bbb Z^+$ let $x_n\ne 0$ with $\|x_n\|_1>n\|x_n\|_2.$ Then let $y_n=x_n/ (\|x_n\|_2\cdot \sqrt n).$ But then $\|y_n\|_2\to 0$ and $\|y_n\|_1>\sqrt n,$ contrary to $\|y_n-0\|_2\to 0\implies \|y_n-0\|_1\to 0.$