I have become really interested in trying to prove directly things that are more easily proved by contradiction. The below seems to be a good example of this. Rudin's proof makes perfect sense to me. In fact I think it is a very elegant proof. But nonetheless I am very much interested in a constructive proof.
Essentially Rudin proves that every k-cell is compact by assuming the opposite: that there is an open cover of a k-cell $I$ which does not contain a finite subcover. I would like to assume $\{G_\alpha \}$ is an arbitrary open cover of $I$ and show directly that $G_\alpha$ must have a finite subcover.
Here is an outline of Rudin's proof(Theorem 2.40):
Every k-cell is compact.
Proof. Let $I$ be a k-cell consisting of all points $x = (x_1, \dots, x_k)$ such that $a_j \leq x_j \leq b_j$ for $1 \leq j \leq k$. Put
$\delta = \{ \sum\limits_1^k(b_j – a_j)^2\}^{1/2}$
Then $|x-y| \leq \delta$ if $x, y \in I$.
Suppose there is an open cover $\{G_\alpha \}$ which contains no finite subcover. Put $c_j = \frac{a_j + b_j}{2}$. The intervals $[a_j, c_j]$ and $[c_j, b_j]$ determine $2^k$ k-cells whose union is $I$. At least one of these subsets of $I$, say $I_1$, cannot be covered by any finite subcollection of $\{ G_\alpha \}$. So we begin again with the k-cell $I_1$ and subdivide further to achieve a sequence of k-cells such that
(a) $I \supset I_1 \supset I_2 \supset I_3 \supset \dots$
(b) $I_n$ is not covered by any finite subcollection of $G_\alpha$
(c) If $x \in I_n$ and $y \in I_n$ then $|y-x| \leq 2^{-n}\delta$
Hence there is a point $x^* \in \cap I_n$ and for some $\alpha$ $x^* \in G_\alpha$. Since $G$ is open there is a neighborhood $N_r(x^*) \subset G_\alpha$. If $n$ is large enough that $2^{-n}\delta < r$ then given $p \in I_n$ $|x^* – p | < 2^{-n}\delta < r \implies p \in N_r(x^*) \implies I_n \subset G_\alpha$ contradicting the fact that $I_n$ cannot be covered by a finite subcollection of $\{ G_\alpha\}$
Best Answer
https://personal.math.ubc.ca/~malabika/teaching/ubc/fall18/math320/HW7-revised.pdf
This assignment provides an alternative way to prove it. Is this what you want?