Let $V$ be a vector space and $\lVert\cdot\rVert_a$, $\lVert\cdot\rVert_b$ be two norms on this space. Let $A=\{x\in V\mid\lVert x\rVert_a\leq 1\}$ and $B=\{x\in V\mid\lVert x\rVert_b\leq 1\}$ be the closed ball with center $\mathbb{0}$ and radius $1$ in $\lVert\cdot\rVert_a$, $\lVert\cdot\rVert_b$ respectively.
I want to show that, if $A$ is bounded in $(V,\lVert\cdot\rVert_b)$ and $B$ is bounded in $(V,\lVert\cdot\rVert_a)$, then $\lVert\cdot\rVert_a$ and $\lVert\cdot\rVert_b$ are equivalent, i.e. $\exists c,c'>0:c\lVert x\rVert_a\leq \lVert x\rVert_b\leq c'\lVert x\rVert_a$.
What I've tried so far: Since the balls are bounded in the respective opposite space, there exists a radius and a center point such that they are contained in an open ball in the respective space. Thus, we have
$$ \lVert x\rVert_a\leq 1\text{ implies }\lVert x_0-x\rVert_b< r $$
and
$$ \lVert x\rVert_b\leq 1\text{ implies }\lVert x_0'-x\rVert_a< r' $$ I'm not sure how I should proceed from this.
Since $A$ is bounded in $\bigl(V,\|\cdot\|_b\bigr)$, there is a $r>0$ such that $A\subset\left\{v\in V\,\middle|\,\|v\|_b\leqslant r\right\}$. By the same reason, there is a $r'>0$ such that $B\subset\left\{v\in V\,\middle|\,\|v\|_a\leqslant r'\right\}$.
Now, take $v\in V\setminus\{0\}$. Then $\left\|\frac v{\|v\|_a}\right\|_a=1$ and therefore $\left\|\frac v{\|v\|_a}\right\|_b\leqslant r$. In other words, $\|v\|_b\leqslant r\|v\|_a$ and this is obviously true if $v=0$. By the same argument, $\|v\|_a\leqslant r'\|v\|_b$. Therefore, the norms are equivalent.