I'm reading a proof of the following claim. The proof contains some claims I am not following, can someone elaborate?

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> *Claim:*
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> Denote the closed unit balls in $X$ and $X^\*$ by $B$ and $B^\*$, respectively. Denote $\mathcal{F}(B)$ to be a set of functions of the form $f:B \rightarrow [-1,1]$ with pointwise product topology (for basis, see below). Define the restriction map $R: B^\* \rightarrow \mathcal{F}(B)$ by $R(\phi) = \phi|\_B$ for $\phi \in B^\*$. Show $R(B^\*)$ is a closed subset of $\mathcal{F}(B)$.
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> *Proof:*
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> Let $f: B \rightarrow [-1,1]$ be a point of closure with respect to the pointwise product topology of $R(B^\*)$ (defn of topology given below). To show $f \in R(B^\*)$ it suffices to show that for all $u, v \in B$ and $\lambda \in \mathbb{R}$ for which $u + v$ and $\lambda u$ also belong to $B$,
> $$f(u+v)= f(u)+f(v) \text{ and } f(\lambda u)= \lambda f(u)$$
> However, for any $\epsilon > 0$, the weak-\* neighborhood of $f$, $\mathcal{N}\_{\epsilon, u,v,u+v}(f)$ (see below for definition, this is an element of the basis of weak-\* topology), contains some $R(\phi\_\epsilon)$ and since $\phi\_{\epsilon}$ is linear, we have $|f(u + v) - f (u) - f (v)| < 3\epsilon$. Therefore, $f(u+v)= f(u)+f(v)$ holds. The proof of $f(\lambda u)= \lambda f(u)$ is similar.
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> In the proof above, we denote a basis element of the weak-\* topology as: $$\mathcal{N}\_{\epsilon,x\_1,...x\_n}(\psi) =\bigg \{\psi' \in B^\* \, \bigg| \, |\psi'(x\_i)-\psi(x\_i) | < \epsilon \quad \forall i = 1,...n\bigg\} \\ =\bigg \{\psi' \in \mathcal{F}(B) \, \bigg| \, |\psi'(x\_i)-\psi(x\_i) | < \epsilon \quad \forall i = 1,...n\bigg\}$$
> The pointwise product topology (the basis shown in the second equality) and weak-\* topology are homeomorphic (there's a bijection between basis elements established by the equality above).
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Questions:

1. *"To show $f \in R(B^\*)$ it suffices to show..."* -- What are they verifying about $f$ here? Loosely, my understanding is that proofs of closure proceed as follows: suppose $x \in \overline{X}$, check neighborhoods of $x$. In particular, we suppose $x$ exists, there are no "properties" of $x$ to check.
2. "*for any $\epsilon > 0$, the weak-$^\*$ neighborhood of $f$, $\mathcal{N}\_{\epsilon, u,v,u+v}(f)$ ... contains some $R(\phi\_\epsilon)$*" -- What is $\phi\_{\epsilon}$? Why does there exist $R(\phi\_\epsilon) \in \mathcal{N}\_{\epsilon, u,v,u+v}(f)$? This is where they are establishing that the $f$ they chose is a limit point, right?
3. "*However...* [to the end]" -- What's being verified here? Aren't we done after we show every neighborhood contains a point of the set? (i.e. after step 2).