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Few questions about the theorem

If $V$ is a neighborhood of $0$ in a topological vector space $X$ and if
$$
K = left{lambda in X^* : |Lambda x | leq 1 ; text{for every} ; x in V right}
$$
then $K$ is weak* compact.

I'll comment on the proof

Proof: Since neighborhoods of $0$ are abosrbing, there corresponds to each $x in X$ a number $gamma(x) < infty$ such that $x in gamma(x)V$. Hence
$$
|Lambda x|leq gamma(x) ; (x in X, Lambda in K) ;;;;(1)
$$
Let $D_x$ be the set of all scalars $alpha$ such that $|alpha| leq gamma(x)$. Let $tau$ be the product topology on $P$, the cartesian product of all $D_x$, one for each $x in X$. Since each $D_x$ is compact, so is $P$, by Tychinoff's theorem. The elements of $P$ are the functions $f$ on $X$ (linear or not) that satisfy
$$
|f(x)| leq gamma(x) ;;; (x in X) ;;;;; (2).
$$

Why are such $f$ the elements of $P = prod_{x in X} D_x$?

Thus $K subset X^* cap P$. It follows that $K$ inherits two topologies, one from $X^*$ (it's weak* topology, to which the conclusion of the theorem refer) and the other, $tau$, from $P$. We will see that

(a) these two topologies coincide on $K$, and

(b) $K$ is a closed subset of $P$

Since $P$ is compact, (b) implies that $K$ is $tau$ compact, and then (a) implies that $K$ is weak*-compact.
Fix $Lambda_0 in K$. Choose $x_i in X, 1 leq i leq n$; choose $delta > 0$.
Put
$$
W_1 = left{Lambda in X^* : |Lambda x_i - Lambda_0 x_i | < delta ; for ; 1 leq i leq n right} ;;;; (3)
$$
and
$$
W_2 = left{f in P : |f(x_i) - Lambda_0 x_i | < delta ; for ; 1 leq i leq n right} ;;;; (4)
$$
Let $n, x_i$ and $delta$ range over all admissible values. The resulting sets $W_1$ form a local base for the weak* topology of $X^*$ at $Lambda_0$ and the sets $W_2$ form a local base for the product topology of $P$ at $Lambda_0$. Since $K subset P cap X^*$, we have
$$
W_1 cap K = W_2 cap K.
$$
This proves (a).

Why do we have both
$$
K subset P cap X^*
$$
and
$$
W_1 cap K = W_2 cap K.
$$
?

The rest of the theorem is clear to me.

Let $F={f:Xrightarrow mathbb{R}: |f(x)|leq gamma(x)}$. $P=prod_{xin X} D_x$ this implies that an element of $P$ is a family $(a_x)_{xin X}$ where $a_xin D_x$, which is equivalent to saying that $|a_x|leq gamma(x)$. Write $f(x)=a_x$ then $f(x)in F$.

Conversely, if you have $fin F$, $|f(x)|leq gamma(x)$ implies that $f(x)in D_x$, so the map $Frightarrow prod_{xin X}D_x$ which assigns $(f(x))_{xin X}$ is well defined and is a bijection.

$K={lambdain X^*:|lambda(x)|leq 1, xin V}$. Let $yin X$, there exists $gamma(y)$ and $zin V$ with $y=gamma(y)z$, we have $|f(y)|=|gamma(y)||f(z)|leq |gamma(y)|$ since $|f(z)|leq 1$, we deduce that $(f(y))_{yin X}in P$, this implies that $fin P$.

$W_1={Lambdain X^*:|Lambda x_i-Lambda x_0|leq delta, 1leq ileq n}$, this implies that $W_1cap K={Lambdain X^*cap K:|Lambda x_i-Lambda x_0|leq delta, 1leq ileq n}={Lambdain K:|Lambda x_i-Lambda x_0|leq delta, 1leq ileq n}$. since $Ksubset X^*cap Psubset X^*$.

$W_2={Lambdain P^*:|Lambda x_i-Lambda x_0|leq delta, 1leq ileq n}$, this implies that $W_2cap K={Lambdain Pcap K:|Lambda x_i-Lambda x_0|leq delta, 1leq ileq n}={Lambdain K:|Lambda x_i-Lambda x_0|leq delta, 1leq ileq n}$ since $Ksubset X^*cap Psubset P$.

We deduce that $W_1cap P=W_2cap P$.