On Rudin there's the following definition:
A local base of a topological vector space $X$ is thus a collection $B$ of neighborhoods of $0$ such that every neighborhood of $0$ contains a member of $B$
This definition makes no sense for me. For example, if we replace member of $B$ by 'neighborhood of $0$' then the above definition becomes:
A local base of a topological vector space $X$ is thus a collection $B$ of neighborhoods of $0$ such that every neighborhood of $0$ contains a neighborhood of $0$
No, it's a collection $\mathcal{B}$ of neighbourhoods of $0$ such that if $O$ is any neighourhood of $0$ there exists some $B \in \mathcal{B}$ such that $B \subseteq O$.
A typical example in $\mathbb{R}$ is $\mathcal{B} = \{(-\frac{1}{n}, \frac{1}{n}): n \in \mathbb{N}^+\}$. These open intervals are all open neighbourhoods of $0$ and every neighbourhood $O$ of $0$ contains some ball of radius $r>0$ around $0$, so $(-r,r) \subseteq O$, and then finding $k$ with $\frac{1}{k} < r$, we see that we can take $(-\frac1k, \frac1k) \in \mathcal{B}$ and $(-\frac1k, \frac1k) \subseteq (-r,r) \subseteq O$.
The point is that sometimes we can take a small (here countable) set of neighbourhoods that are all we need for considering continuity at $0$, or closure properties of $0$, instead of the full set of neighbourhoods.