Everything is in the context of metric spaces.
My definition of compactness is: every sequence has a convergent subsequence.
Let $B_1, B_2 ...$ be a countable open cover of a compact set $A$. Suppose there is no finite subcover. Then, $\exists$ $a_1 \in A \setminus B_1$,
$\exists$ $a_2 \in A \setminus (B_1 \cup B_2)$,
$\exists$ $a_3 \in A \setminus (B_1 \cup B_2 \cup B_3)$,
$\exists$ $a_n \in A \setminus (B_1 \cup \cdots \cup B_n)$,
But $A$ is compact, so $\exists$ a subsequence $a_{n_k}$ such that $\lim \limits_{k \to \infty} a_{n_k} = a \in A$.
there exists $n_0$ such that $a \in B_{n_0}$ so all but finitely many elements in the subsequence $a_{n_k}$ are inside $B_{n_0}$.
Please help me complete the argument to arrive at a contradiction. I did not understand it in lecture.
I know there are infinitely many $a_{n_k}$'s inside $B_{n_0}$ and only finitely many $a_{n_k}$'s outside $B_{n_0}$.
Note that $a_i \not \in B_j$ if $i > j$. Since $(a_{n_k})_k$ is a subsequence, there exists $k_0$ such that $n_{k_0} > n_0$ and in particular, we have that $n_l > n_0$ for all $l > k_0$. This is absurd since it shows that infinitely many terms of $(a_{n_k})_k$ lie outside of $B_{n_0}$.