Two norms on a linear vector space are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm. I want to prove the following statement:

Two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ on a linear vector space $X$ are equivalent if and only if every set that is bounded in one of the norms is bounded in the other norm.

(Note that going from equivalent to bounded is straightforward, so I don't need help with that.)
As per my book, two norms are equivalent if there exists $\alpha,\beta > 0$ such that
$$
\alpha \| x \|_1 \leq \| x \|_2 \leq \beta \| x \|_1 \hspace{1cm} \forall x \in X
$$
Attempt at solution
Proof by contradiction, i.e. for all $\alpha,\beta > 0$, we have the following inequalites
$$
\| x_0 \|_2 < \alpha\| x_0 \|_1
$$
or
$$
\beta\| x_0 \|_1 < \| x_0 \|_2
$$
for some $x_0$.
I want to use the boundedness assumption here, like, the LHS is bounded by the RHS, thus the RHS is bounded too. But the problem is that this assertion is for $x_0$ specifically, not the entire domain $X$.
Any ideas?
 A: Suppose bounded sets for the two norms are the same. Let $A=\{x: \|x\|_1 \leq 1\}$. This is bounded in the first norm. So it must be bounded in then second norm. Hence ther exists $\beta \in (0,\infty)$ such that $\|x\|_2 \leq \beta$ whenever $\|x\|_1 \leq 1$. Now take any non-zero vector $y$. Let $x=\frac y {\|y\|_1}$. Then $\|x\| \leq 1$ so $\|x\|_2 \leq \beta$. This gives $\|y\|_2 \leq \beta \|y|\|_1$. A similar argument gives $\|y\|_1 \leq c \|y|\|_2$ for some $c>0$ and this can be writen as $\alpha \|y\|_1 \leq \beta \|y|\|_2$ where $\alpha =\frac 1  c$. Hence the norms are equivalent.
Conversely if the norms are equivalent it is evident that boundedness of one the norms for a set of vectors implies boundedness of the set for the other norm.
A: Consider the unit ball in the $\|\cdot\|_1$ norm. It is bounded in the $\|\cdot\|_2$ norm, which means that for any $x$ with $\|x\|_1\le 1$ you have $\|x\|_2\le \beta$ for some $\beta>0$. But since both norms are homogeneous, the same is true for any $x$ in your space. In the same way you prove that for any $x$ with $\|x\|_2\le 1$ you have $\|x\|_1\le \gamma$ for some $\gamma>0$.
