Rudin functional analysis theorem 1.13 (f) and theorem 1.11. Theorem 1.13 (f) states:

Let $X$ a topological vector space. I $E$ is a bounded subset of $X$,
  so is $\bar{E}$.

The proof relies on theorem 1.11, which states without proof:
If $\mathcal{B}$ is a local base for a topological vector space $X$, then every member of $\mathcal{B}$ contains the closure of some member of $\mathcal{B}$.
Proof of theorem 1.13 (f) proceeds as follow:
Let $V$ a neighborhood of $0$. By theorem 1.11 we have $\bar{W} \subset V$ for some neighborhood of $W$ of $0$. Since $E$ is bounded we have $E \subset tW$ for $t$ large enough. For these $t$, we have $\bar{E} \subset t \bar{W} \subset tV$ so $\bar{E}$ is bounded.
Now the questions are two:


*

*Why is theorem 1.11 true?

*Why applying such theorem we find $\bar{W} \subset V$ for some open $W$?


Update: I'm trying to use Rudin convention to prove theorem 1.11, I assume I should use theorem 1.13
If $K = \left\{ 0\right\}$ and $C$ closed then there's a neighbour $V$ of $0$ such that
$$
V \cap (C + V) = \emptyset
$$
by definition of local base $\mathcal{B}$ there's a $\gamma \in \mathcal{B}$ such that $\gamma \subset V$ and therefore
$$
\gamma \cap (C + \gamma) = \emptyset,
$$
but $C$ is closed and and $Int(C) \subset C$ therefore $Int(C) + \gamma$ is a subset of $C + \gamma$, $Int(C)$. If I assume $0 \in C$ then $Int(C) + \gamma$ is a neighbour of $0$, then there's a member $\gamma' \in\mathcal{B}$ such that $\gamma' \subset Int(C) + \gamma$.
Is all the above correct? I guess if it is then I can somehow prove that $\bar{\gamma'} \subset \gamma$, any hint?
 A: Rudin uses the convention that neighbourhoods are by definition open sets. If we use the (in my opinion more convenient) convention that does not require neighbourhoods to be open, we can state theorem 1.11 as

The family of closed neighbourhoods of a point $p$ is a local base at $p$.

In other words, a topological vector space is a $T_3$ space. Since Rudin requires a topological vector space to be $T_1$ (singleton sets are closed), this is true regardless of which nomenclature with respect to regular/$T_3$ (there are two widespread nomenclatures, in one we have "regular = $T_3 + T_0$", in the other "$T_3 = {}$ regular ${} + T_0$") one follows.
This follows easily from the continuity of the vector space operations. For $p \in X$ let $\mathscr{V}(p)$ be the family of neighbourhoods of $p$. Let $A \subset X$ and $V \in \mathscr{V}(0)$. Then
$$A + V = \{ a + v : a \in A, v \in V\} = \bigcup_{a \in A} (a + V)$$
is called a uniform neighbourhood of $A$. In topological vector spaces we have
$$\overline{A} = \bigcap_{V \in \mathscr{V}(0)} (A + V)\,,$$
i.e. the closure of a set is the intersection of all uniform neighbourhoods of the set. To see this, fix an arbitrary $V \in \mathscr{V}(0)$ and note
$$p \in A + V \iff \bigl(\exists a \in A\bigr)\bigl(p \in a + V\bigr) \iff \bigl(\exists a \in A\bigr)\bigl(a \in p - V\bigr) \iff (p - V)\cap A \neq \varnothing\,.$$
Since the translation $\tau_p \colon x \mapsto p + x$ and multiplication with $-1$ are homeomorphisms, $p - V$ runs through $\mathscr{V}(p)$ when $V$ runs through $\mathscr{V}(0)$ and thus
\begin{align}
p \in \bigcap_{V \in \mathscr{V}(0)} (A + V)
&\iff \bigl(\forall V \in \mathscr{V}(0)\bigr)\bigl(p \in A + V\bigr) \\
&\iff \bigl(\forall V \in \mathscr{V}(0)\bigr)\bigl((p - V)\cap A \neq \varnothing\bigr) \\
&\iff \bigl(\forall W \in \mathscr{V}(p)\bigr)\bigl(W \cap A \neq \varnothing\bigr) \\
&\iff p \in \overline{A}\,.
\end{align}
Now let $p \in X$ and $V \in \mathscr{V}(p)$. By the continuity of addition at the point $(p,0) \in X\times X$ there is a neighbourhood $W_0$ of $(p,0)$ such that $(x,y) \in W_0 \implies x+y \in V$. By definition of the product topology, there is a neighbourhood $W_1$ of $p$ and a neighbourhood $U$ of $0$ with $W_1 \times U \subset W_0$, and hence we have
$$\overline{W}_1 \subset W_1 + U \subset V\,.$$
To complete the proof of theorem 1.11 as stated by Rudin, assume $V \in \mathcal{B}$, and note that since $\mathcal{B}$ is a local base (at $p$) there is a $W \in \mathcal{B}$ with $W \subset W_1$ (where $W_1$ is as above) and hence $\overline{W} \subset \overline{W}_1 \subset V$.
Now in the proof of 1.13 (f) we choose $\mathcal{B} = \mathscr{V}(0)$. We could choose any other neighbourhood basis of $0$, but since we don't need any special property of the local base, we might as well take the conceptually simplest. And in that case, theorem 1.11 says "every neighbourhood $V$ of $0$ contains the closure of some other neighbourhood $W$ of $0$". In symbols,
$$\bigl(\forall V \in \mathscr{V}(0)\bigr)\bigl(\exists W \in \mathscr{V}(0)\bigr)\bigl(\overline{W} \subset V\bigr)\,.$$
And then, since $E$ is assumed to be bounded there is a $t > 0$ with $E \subset tW$, from which it follows that
$$\overline{E} \subset \overline{tW} = t\overline{W} \subset tV\,.$$
Since $V \in \mathscr{V}(0)$ was arbitrary, it follows that $\overline{E}$ is bounded.
