Two questions about a proof of the compactness of an operator There are a few things that I don't understand about a proof and I'd appreciate any help. The theorem and its proof are the following:

(1) Is the equality 
$$ \|v(\tau) -v(\tau_j)\| = \max_{1 \le i \le n}|u^\ast (\tau) (x_i) - u^\ast(\tau_j) (x_i)|$$
a typo? Shouldn't it be
$$ \|v(\tau) -v(\tau_j)\| = \max_{s \in S}|u^\ast (\tau) (s) - u^\ast(\tau_j) (s)|$$ 
? Obviously, the latter implies the former. 
(2) To prove the claim all we need to do is prove that each term in the inequality 
$$ \begin{align}
&|u^\ast(\tau)(x) - u^\ast(\tau_j)(x)| \le \\
&|u^\ast (\tau) (x) -u^\ast (\tau) (x_i)| + |u^\ast(\tau)(x_i)-u^\ast(\tau_j)(x_i)| + |u^\ast (\tau_j) (x_i) -u^\ast (\tau_j)(x)| 
\end{align}$$
is less than $\varepsilon / 3$. The middle term is clear. But as far as I can tell the other two terms are not mentioned in this proof. So I was wondering if one can argue as follows (the same argument should work for both):
$\tau$ is in the closed unit ball of $Y^\ast$ and by definition $u^\ast (\tau) = \tau \circ u$. Since $\tau$ has finite rank it is compact. Also, $u$ is compact by assumption and the composition of compact operators is compact. Hence $\tau \circ u (S)$ is totally bounded. Hence there exist $x_1, \dots, x_n$ in $S$ such that $\|u^\ast (\tau) (x) - u^\ast (\tau)(x_i)\| < \varepsilon / 3$ for some $i$. 
Basically, I think at the beginning where the author talks about $u$ they really wanted to talk about $u^\ast (\tau)$. Can someone confirm this or clarify why my thoughts are wrong?
 A: For (1), look at how $v(\tau)$ is defined.  Recall
$$
v(\tau) = (\tau u(x_1), \tau u(x_2), \cdots, \tau u(x_n)),
$$
which is an element of $\mathbb{C}^n$.  Using the sup-norm on $\mathbb{C}^n$ and the definition of $v(\tau)$ gives
$$
\|v(\tau) - v(\tau_j)\| = \max_{1\leq i \leq n} |u^*(\tau)(x_i) - u^*(\tau_j)(x_i)|.
$$
For (2), the first term is less than $\varepsilon / 3$ because of the total boundedness of $u(S)$.  Look at the second sentence: "there exist elements $x_1, x_2, \cdots, x_n$ such that if $x\in S$, then $\|u(x) - u(x_i)\| < \varepsilon / 3$ for some $1\leq i\leq n$".  So choosing the right $i$ makes the first term small.
The smallness of the third term also from  $\|u(x) - u(x_i)\| < \varepsilon / 3$, as well as $\|\tau\| \leq 1$.  (Recall $\tau \in T$, and $T$ was the unit ball).  So 
$$
|u^*(\tau_j)(x_i) - u^*(\tau_j)(x)| \leq \|\tau\|\|u(x_i) - u(x)\| 
< \varepsilon / 3.
$$
This was mentioned in the proof (without the explicit calculation) in the sentence right before the word "Hence".
Hope this helps!
