Exercise: Let $B$ be an open bounded neighborhood of $0$ in a topological vector space. Show that every neighborhood of $0$ contains a set of the form $\{sB : s\in(0,\infty)\}$.
Hint: One can use the fact, that every neighborhood of $0$ is absorbent and that the set $B$ is bounded.
Proof: $B$ is bounded implies the following: For any $V$, a neighborhood of $0$, there is a $s>0$, such that $B\subset sV$. Then clearly $s^{-1}B$ is again a neighborhood of $0$, since multiplication with $s^{-1}$ is a homeomorphism, $s^{-1}B\subseteq V$ and $s^{-1}>0$.
Questions: Is this proof right? Where do I need the fact, that every neighborhood of $0$ is absorbent? Doesn't this even turn $\{sB : s\in(0,\infty)\}$ into a neighborhood basis of $0$?
Any hint or help is appreciated! Thank you in advance!