I am trying to prove the Heine Borel theorem for compactness of the closed interval $[0,1]$ using Konig's lemma. This is what I have so far:
I assume $[0,1]$ can be covered by $\{(a_i,b_i):i=0,1,2\cdots\}$.
I construct a graph $G$ as follows: First let a vertex be labelled $[0,1]$ (the root). Then consider $[0,1]-(a_0,b_0)\cup(a_1,b_1)$. This consists of $n_1$ closed intervals where $n_1$ is finite. Adjoin the $[0,1]$ vertex with $n_1$ vertices labelled by these closed intervals (these vertices will be at level 1). Next consider $[0,1]-(a_0,b_0)\cup(a_1,b_1)\cup(a_2,b_2)$. This consists of $n_2$ closed intervals. Each of these closed intervals is a subset of exactly one the closed intervals considered in the previous step. Make $n_2$ vertices labelled by these closed intervals and adjoin them to that vertex created in the previous step of which it is a subset of (these vertices will be at level 2). Continue doing so for higher levels, each time obtaining the labels by considering the closed interval obtained from $[0,1]-(a_0,b_0)\cup(a_1,b_1)\cdots \cup(a_i,b_i)$.
This yields a rooted tree $G$ where each level is finite.
Suppose the tree contained an infinite path: $[0,1]\supset[\alpha_1,\beta_1]\supset[\alpha_2,\beta_2]\cdots$.
Since a sequence of nested closed intervals is nonempty so there is an element $x$ in it. As $x\in [0,1]$ so $x\in(a_i,b_i)$ for some $i$. But then $x$ cannot exist in any interval which is at a level beyond $i$, yielding a contradiction to 4.
So by the contrapositive form of Konig's lemma, $G$ cannot be infinite. It follows that for some $i$, $[0,1]-(a_0,b_0)\cup(a_1,b_1)\cdots \cup(a_i,b_i)$ is empty. Hence $[0,1]$ is covered by $(a_0,b_0)\cup(a_1,b_1)\cdots \cup(a_i,b_i)$.
My doubts are in the arguments presented in 2. and 6. Are they correct? In particular is this statement: "Each of these closed intervals is a subset of exactly one the closed intervals considered in the previous step." correct?
What is an upper bound for $n_k$?
Thanks