The dual of $w^{\star}$ topological space There is a result in my book states:

Suppose that $(X, \tau)$ is a locally convex topological vector spaces. $(X, \tau)^{\star}$ is the dual of $(X, \tau)$ and denoted by $X^{\star}$. Let $\sigma(X^{\star}, X)$ be the $w^{\star}$ topology on $X^{\star}$, then $(X^{\star}, \sigma(X^{\star}, X))^{\star} = (X, \tau)$.

There has been a question related to mine ( see dual space of a locally convex space ), however, it is not related to the seeking of a proof of the result $(X^{\star}, \sigma(X^{\star}, X))^{\star} = (X, \tau)$ and the answer to it is also not helpful for me.  Thus I raised my question here.
In fact, the proof for $(X, \tau) \subseteq (X^{\star}, \sigma(X^{\star}, X))^{\star}$ is an easy job. For me, the hard pary is to prove the reversed inclusion relation $(X, \tau) \supseteq (X^{\star}, \sigma(X^{\star}, X))^{\star}$. 
Of course, there is a proof in my book ( in Chinese, and omit its title ). But I have some troubles in understanding it. Let me rewrite it below:
The proof: 
If $y \in (X^{\star}, \sigma(X^{\star}, X))^{\star}$, then for any $\varepsilon>0$, there exists a neighbourhood of $0$, denoted by $U(x_1, \cdots, x_n ; \delta)$, such that $U(x_1, \cdots, x_n ; \delta) \subseteq y^{-1}(-\varepsilon, \varepsilon)=\{ x^{\star}: |y(x^{\star})|<\varepsilon \}$, where 
$$U(x_1, \cdots, x_n ; \delta)\overset{\text{def}}{=}\{ x^{\star}: |x_{i}(x^{\star})|<\varepsilon, i=1, \cdots, n \}, ~ x_i \in X, ~ n\in \mathbb{N}.$$ 
Thus, $x^{\perp}_{1} \bigcap \cdots \bigcap x^{\perp}_{n} \subseteq y^{\perp}$, where $x^{\perp} \overset{\text{def}}{=} \{ x^{\star}: x(x^{\star})=0 \}$. Hence, $y=\sum^n_{i=1} \alpha_{i}x_{i}$, for a appropriate set of real numbers $(\alpha_1, \cdots, \alpha_n).$ Therefore $y \in X$.
My questions:
$\bf{1}$. Since $\{x_i\}$ are related to the choice of $\varepsilon$, how can we get $x^{\perp}_{1} \bigcap \cdots \bigcap x^{\perp}_{n} \subseteq y^{\perp}$ ? However, it may be wrong if we start like this: $\forall z^{\star} \in x^{\perp}_{1} \bigcap \cdots \bigcap x^{\perp}_{n}$, for $\forall \varepsilon>0$, there is an open neighbourhood $U(x_1, \cdots, x_n ; \delta)$ such that $U(x_1, \cdots, x_n ; \delta) \subseteq \{ x^{\star}: |y(x^{\star})|<\varepsilon \}\dots$ Unless $x_1, \cdots, x_n$ can be fixed with the choice of $\varepsilon$, why ?
$\bf{2}$. How the result $y=\sum^n_{i=1} \alpha_{i}x_{i}$ is derived in the proof ? However, I can't see any clues in it.
Thanks in advance.
 A: For the first question, if for a fixed $\epsilon > 0$, you can find $\delta > 0$ such that $$U(x_1, \cdots x_n, \delta) \subset y^{-1}(-\epsilon, \epsilon)$$ you can see 
$$U(x_1, \cdots x_n, \dfrac{\delta}{k}) \subset y^{-1}(-\dfrac{\epsilon}{k}, \dfrac{\epsilon}{k}), \forall k \in \mathbb{N^*}$$
then remark that $$x^{\perp}_{1} \bigcap \cdots \bigcap x^{\perp}_{n} = \cap_{k = 1}^{+\infty}U(x_1, \cdots x_n, \dfrac{\delta}{k}) \subset \cap_{k = 1}^{+\infty} y^{-1}(-\dfrac{\epsilon}{k}, \dfrac{\epsilon}{k}) = y^{\perp}$$.
For the second question, lemma 3.9 in Rudin's Fucntional Analysis proves it, I just copy the proof: 
You can define $f(x_1(x^*), \cdots, x_n(x^*)) = y(x^*)$, it is well define since $x^{\perp}_{1} \bigcap \cdots \bigcap x^{\perp}_{n} \subset y^{\perp}$, and $f$ is a linear mapping from $(x_1(X^*), \cdots x_n (X^*))$ to $\mathbb{C}$(or $\mathbb{R}$), you can extend $f$ to a linear mapping on $\mathbb{C}^n$(or $\mathbb{R}^n$), which is $f(c_1, \cdots, c_n) = \sum \alpha_i c_i$, then it's easy to get your conclusion.
