Weierstrass and Heine-Borel theorems in Rudin's 3rd This is a question about the logic of Theorems 2.41, 2.42 in Rudin's 3rd Ed, which  deal with the Heine-Borel and Weierstrass properties of sets of $R^k$, respectively. 
A quick version of my question is: doesn't 2.41 (together with 2.40, on which it depends) moot 2.42? 
2.41. (Heine-Borel+) If a set E in $R^k$ has one of these three properties, it has the other two: (a) E is closed and bounded; (b) E is compact; (c) Every infinite subset of E has a limit point in E. 
2.42. (Weierstrass) Every bounded infinite subset of $R^k$ has a limit point in $R^k$. 
The text explains that 2.41(a) implies (b) implies (c) implies (a). So the only step to be supplied for 2.42 is that a bounded subset of $R^k$ is compact (and therefore closed, to bring it into the ambit of 2.41). 
But in both theorems Rudin resorts to an extrinsic proof (his 2.40) to show that  that "k-cells" are compact.  
I think (am not sure) there is a theorem stating that Heine-Borel implies Weierstrass (and conversely) but H-B consists of 2.41(a) and (b), according to Rudin's note preceding 2.41.  I wonder if, with the addition of 2.41(c), he needs to prove Weierstrass separately?  
This is a sort of fussy question but an answer might help me understand the relationship between these ideas better. Thanks.  EDITED: so the quick version of the question includes the ref. to 2.40.
 A: The connection between the topics is strong, but not as strong as you currently believe.  The problem is that a bounded set of $\mathbb{R}^k$ is not necessarily compact: take the open unitl ball in $\mathbb{R}^k$, or more  generally any bounded set in $\mathbb{R}^k$ that is not closed.  The k-cells are invoked to say that our closed and bounded set $E$ is a subset of some k-cell $I$, as it is bounded.  It would be enough to take $E$'s closure, but this approach also works: as k-cells are compact, 2.41 tells us that every infinite subset of $I$ (which includes the infinite subsets of $E$) has a limit point in $I \subset \mathbb{R}^k$ (and not necessarily $E$).  
Further examining Rudin's setup, both the given proofs for Heine-Borel and Weierstrass use the general observations that infinite subsets of compact sets have a limit point in that compact set, as well as the compactness of k-cells and that closed sets of compact sets are compact. 
A: 2.41 does not trivialize 2.42, as it is not necessarily true that a bounded subset of $\mathbb{R}^k$ is compact. Why? Consider the following set:
$$\left\{\frac{1}{n} | n\in \mathbb{N}\right\}\subset \mathbb{R}$$
This is bounded but not compact. Why? $$\bigcup_{i\in\mathbb{N}} \left(\frac{1}{i+1/2},\frac{1}{i-1/2}\right)$$ is an open cover with no finite subcover. So you really do need both closed and bounded for compact. As a result, you have to supply an extra argument to get 2.42 from 2.41- an outline of such an argument is as follows:
Every bounded infinite subset $A\subset\mathbb{R}^k$ has a closure $A\cup \{\mathrm{limit points of A}\}=\overline{A}\subset\mathbb{R}^k$, which is also a bounded infinite subset of $\mathbb{R}^k$. But $\overline{A}$ is closed, so since it is also bounded, it is compact by 2.41 (a)$\Rightarrow$(b). Then we can apply 2.41 (b)$\Rightarrow$(c) to see that such a set has a limit point in $\overline{A}\subset\mathbb{R}^k$ and thus every bounded infinite subset of $\mathbb{R}^k$ has a limit point in $\mathbb{R}^k$.
