Is this the right way to prove finite cartesian product of compact sets is compact Let $\{(X_i, \mathfrak{T}_i) |  i \in F, F \text{ finite}\}$ be a set of compact topological spaces. Let $\{K_i|i \in F, F \text{ finite}\}$ be a set of compact sets drawn from each $X_i$. 

Then the claim is:
  
  
*
  
*$\prod\limits_{i \in F} X_i$ is compact in the product topology
  
*$\prod\limits_{i \in F} K_i$ is compact in the product topology

I wish to construct a topological proof for claim 2, I am not sure if this is the correct way to go about it, I am mainly not sure about the indices.
My attempt:
Let $\mathcal{U}$ be an open cover of $\prod\limits_{i \in F} K_i$, we wish to show that $\mathcal{U}$ has a finite subcover. 
Let $\mathcal{U}_i = \{U_{ij} | j \in I, I \text{ is arbitrary index set}\}$ be any open cover for $K_i$, then by compactness, it has a finite subcover $\{U_{ij} | i \in F, j \in N \subseteq I, F, N \text{ finite}\}$. 
Then $\{U_{ij} | U_{ij} = \prod\limits_{j \in N} U_{ij}, \forall i \in F\}$ is a finite subcover of $\prod\limits_{i \in F} K_i$.

Does this proof look correct?

It would be nice if someone can tell me why is it that for claim 1 we will need to use the tube lemma. i.e. why can't we do the same as in claim 2
Is there some way to infer claim 2 from claim 1?
 A: No, the argument is not correct (which is why you can’t use it to prove the first result, either). Your covers $\mathscr{U}_i$ of the sets $K_i$ come out of nowhere, without the slightest relationship to the given open cover $\mathscr{U}$ of $\prod_{i\in F}K_i$. Moreover, the members of $\mathscr{U}$ need not be products of the form $\prod_{i\in F}U_i$, where $U_i$ is open in $X_i$: they may be very complicated unions of such sets. For example, in $\Bbb R^2$ the set $\{\langle x,y\rangle\in\Bbb R^2:x^2+y^2<1\}$ is open, but it is not of the form $U\times V$ for any open sets $U,V\subseteq\Bbb R$.
I strongly suggest that you first prove (1). The easiest way to do this is by induction on $n$, the number of factor spaces: the case for $n=2$ is identical to the induction step. For this step can use the tube lemma, if you have it available, or you can give an argument that in effect incorporates the proof of the tube lemma; so long as you work directly with open covers and not with some other equivalent formulation of compactness, you’ll need something along those lines.
You can then prove (2) in essentially the same way.
