There is a part of the proof of the following theorem in Introduction to Real Analysis by William F. Trench which I do not understand.
Theorem $8.2.3$: An infinite subset $T$ of a metric space $A$ is compact if and only if every infinite subset of $T$ has a limit point in $T$.
I understand the proof of the only-if part, so I skip that part and write the proof of the if part.
Let $\mathcal{H}$ denote a covering of $T$ and $N_{\epsilon}(t)$ denote the $\epsilon$-neighborhoud of $t$.
If $\epsilon > 0$, then $T$ can be covered by $\epsilon$-neighborhoods of finitely many points of $T$ (I skip the proof of the preceding sentence). By taking $\epsilon$ successively equal to $1, \frac{1}{2}, \cdots, \frac{1}{n}, \cdots$, we can now conclude that, for each $n$, there are points $t_{1n}, t_{2n}, \cdots , t_{k_nn}$ such that $$T\subset \bigcup_{i=1}^{k_n} N_{\frac{1}{n}}(t_{in}).$$ Denote $B_{in}=N_{\frac{1}{n}}(t_{in})$, $1 \le i \le n$, $n \ge 1$, and define $$\{G_1, G_2, G_3, \cdots\}=\{B_{11}, \cdots, B_{k_11},B_{12}, \cdots, B_{k_22},B_{13}, \cdots, B_{k_33},\cdots\}.$$ If $t \in T$, there is an $H \in \mathcal{H}$ such that $t \in H$. Since $H$ is open, there is an $\epsilon>0$ such that $N_{\epsilon}(t) \subset H$. Since $t \in G_j$ for infinitely many values of $j$ and $lim_{j \to \infty} d(G_j) = 0$, $$G_j \subset N_{\epsilon}(t) \subset H$$ for some $j$. Therefore, if $\{G_{j_i} \}_{i=1}^{\infty}$ is the subsequence of ${G_j}$ such that $G_{j_i}$ is a subset of some $H_i$ in $\mathcal{H}$ (the $H_i$ are not necessarily distinct), then $$T \subset \bigcup_{i=1}^{\infty}H_i.$$ The rest of the proof is to show $T \subset \bigcup_{i=1}^{N}H_i$, which I understand.
My problem is I do not understand why $$T \subset \bigcup_{i=1}^{\infty}H_i.$$
I can say, for each $H \in \mathcal{H}$, there exist infinitely many $G_j$ such that $G_j \subset H$. I can also say there are finitely many $G_j$ that cover $T$. But, I cannot prove there are countably many of $G_j$ that cover $T$, and each of $G_j$ is a subset of some $H \in \mathcal{H}$.
Here is my draft for "if". $\Leftarrow{}$ by contropositive.
We want. $\exists$ a infinite subset E of an infinite subset T has no limit point, then T of A is not compact.
E is not an empty set since E is a infinite set. E has no limit point, then $\forall p \in E$, $\exists N_{\epsilon_p}(p)\cap p = p$.
So let $H = \{N_{\epsilon_p}(p)|p\in E\}$. H is an opencover of set E and we can not find a finite subcover of H, then H is not compact.
let set $K = \cup_{p}H\cup \complement_TE$, and K is a opencover of T, we also can't find a finite subcover of K.
so T is not compact.