Clarification: Every infinite subset of E has a limit point of E iff E is compact.

Question

Every infinite subset of E has a limit point of E iff E is compact.

Is this always true?

Answer 1

The statement in the title is false in general: a topological space $X$ that satisfies the condition

(*) Every infinite subspace $E$ of $X$ has a limit point.

is called limit point compact. And a product of two spaces that are limit point compact need not be limit point compact.

This shows that we cannot rely only on this property (*) to show that the product of two compact sets of reals is compact. We need to use more.

E.g. one can use that for metric spaces (like the reals) property (*), compactness and

(**) Every sequence in $X$ has a convergent subsequence.

(which is called sequential compactness) are equivalent. It is quite straightforward to show that the product of two sequentially compact spaces is again sequentially compact (and thus compact, by the equivalence).

For compact subsets of the reals we can also use the compact iff closed and bounded equivalence (which holds for Euclidean spaces, but not for all metric spaces, so it's a less general proof). The ease of this proof will depend on whether you already know that the product of closed sets is again closed.

Of course, any product of compact sets is compact by Tychonoff's theorem, which is overkill here, but the finite case is not too hard, using the open cover definition.