You are right that $G$ is already a finite subcover of itself, so that isn't particularly useful. Perhaps a more interesting characterization of compactness is that every infinite cover contains a finite subcover. (Do you see why they are equivalent?)
Your second example cover, $$\left\{\left(\frac1n,1\right):n\geq 1\right\},$$ is an example of an infinite cover from which no finite subcover can be extracted. This shows that $(0,1)$ is not compact.
As you said, $\phi_t$ is a bounded linear funcional, so as
$$V_t = \phi_t^{-1} \big( \{ |x| > \frac 13 \}\big), $$
then $V_t$ is open in the weak topology. Similarly $U_t$ is also open. It is quite obvious that
$$\{V_t:t\in \mathbb{Q}\cap[0,1]\} \cup \{U_t:t\in (\mathbb{R} \setminus \mathbb{Q})\cap[0,1]\}$$
is an open cover of the unit ball in $C[0,1]$ (Actually, it is an open cover of the whole $C[0,1]$).
Now we show that: for any finite set $t_1, \cdots t_m \in \mathbb Q\cap [0,1]$, $s_1, \cdots s_n \in \mathbb R\setminus \mathbb Q \cap [0,1]$, there is a function $f \in C[0,1]$ such that $||f||\leq 1$ and
$$f\notin \bigcup_{j} V_{t_j} \cup \bigcup_{k} U_{s_k}$$
Well, it just say that I can find a continuous function $f$ such that $|f(t_j)|\leq 1/3$ (so not in $V_{t_j}$) and $|f(s_k)| \geq 2/3$ (so not in $U_{s_k}$) and $||f|| \leq 1$. Such a function can be found by setting
$$f(t_j) =1/3, \ \ \ \ f(s_k) = 2/3$$
and joining line along adjacent $t_j$, $s_k$. This shows that we cannot pick a finite subcover to cover the closed unit ball in $C[0,1]$, thus the unit ball is not weakly compact.
Best Answer
Because the only compact are finite sets. For example, let take the covering $\{\{x\}\}_{x\in B(0,1)}$ can you find a finite sub-covering of $B(0,1)$ ?