Arzela Ascoli, help to understand some points in the proof. Hi everyone I'd like if someone could give me an explanation of some points in the following proof, explicitly the points  with the asterisk. This is from Dudley's, one direction is completely easy, only I put what I have problems to understand. Thanks in advance for your help :)

(Arzela Ascoli) Let $(K,e)$  be a compact metric space and $\mathcal{F}\subset C(K)$ (where $C(K)$ are the continuous functions from $K$ to $\mathbb{R}$). Then $\mathcal{F}$ is totally bounded by $d_{sup}$ iff it is uniformly bounded and equicontinuous.

($\Leftarrow$) Let $\mathcal{F}$ be uniformly bounded and equicontinuous, hence uniformly equicontinuous. Let $|f(x)|\le M<\infty$ for all $f\in \mathcal{F}$ and $x\in K$. Then $[-M,M]$ is compact. Let $\mathcal{G}$ be the closure of $\mathcal{F}$ in the product topology $\mathbb{R}^K$. Then $\mathcal{G}$ is compact by the Tychonoff's theorem $(*)$. For any $\varepsilon>0$ and $x,y\in K$, $\{f\in \mathbb{R}^K: |f(x)-f(y)|\le \varepsilon\}$ is closed $(*^2)$. So if $e(x,y)<\delta$ implies $ |f(x)-f(y)|\le \varepsilon$ for all $f\in \mathcal{F}$, the same remains true for all $f\in \mathcal{G}$ $(*^3)$. Thus $\mathcal{G}$ is also uniformly equicontinuous.
Let $\mathcal{U}$ any ultrafilter in $\mathcal{G}$. Then $\mathcal{U}$ converges for the product topology to some $g\in \mathcal{G}$. Given $\varepsilon>0$, take $\delta>0$ such that whenever $e(x,y)<\delta,\,|f(x)-f(y)|\le \varepsilon/4$ for all $f\in \mathcal{G}$. Take a finite set $S\subset K$ such that for any $y\in K$, $e(x,y)<\delta$ for some $x\in S$. Let 
$$U=\{f: |f(x)-g(x)|< \varepsilon/3 \text{ for all } x\in S\}$$
Then $U$ is open in $\mathbb{R}^K$ $(*^4)$, so $U\in \mathcal{U}$. If $f\in U$, then $|f(y)-g(y)|< \varepsilon$ for all $y\in K$, so $d_{sup}(f,g)\le \varepsilon$. Thus $\mathcal{U}\to g$ for $d_{sup}$. So $\mathcal{G}$ is compact for $d_{sup}$, hence $\mathcal{F}$ is totally bounded.
For $(*)$ I think that is simple because $\mathcal{G}\subset [-M,M]^K$  the latter is compact and the former is closed. But for the other I have troubles to visualize the closed and open sets.
 A: For $(\ast)$, you think correctly. $[-M,M]^K$ is compact by Tíkhonov's theorem, and since $\mathcal{G}$ is by definition a closed subset of that, it is compact too.
For $(\ast^2)$: The map $\eta_{x,y} \colon f\mapsto (f(x),f(y))$ from $[-M,M]^K \to \mathbb{R}^2$ is continuous, hence
$$\{ f \in [-M,M]^K : \lvert f(x) - f(y)\rvert \leqslant \varepsilon\} = \eta_{x,y}^{-1}(\{(u,v)\in\mathbb{R}^2 : \lvert u-v\rvert\leqslant\varepsilon\})$$
is closed in $[-M,M]^K$ as the preimage of a closed set under a continuous map for any two $x,y\in K$.
For $(\ast^3)$: The intersection of closed sets is again closed, so
$$C(\varepsilon,\delta) := \bigcap_{e(x,y) < \delta} \eta_{x,y}^{-1}(\{(u,v)\in\mathbb{R}^2 : \lvert u-v\rvert\leqslant\varepsilon\})$$
is a closed subset of $[-M,M]^K$ for any two $\delta,\varepsilon > 0$. Now, since $\mathcal{F}$ is equicontinuous, for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\mathcal{F} \subset C(\varepsilon,\delta)$. Hence also $\mathcal{G} = \overline{\mathcal{F}} \subset C(\varepsilon,\delta)$. And that means $\mathcal{G}$ is (uniformly) equicontinuous.
For $(\ast^4)$: $U$ is the intersection of finitely many sets of the form $\pi_x^{-1}\bigl((g(x)-\varepsilon/3,g(x)+\varepsilon/3)\bigr)$, where $\pi_x \colon f \mapsto f(x)$ is a coordinate projection. The coordinate projections are continuous, hence these sets are open, and finite intersections of open sets are open. So $U$ is open in the product topology. Hence it is a neighbourhood of $g$, and therefore belongs to all filters converging to $g$.
