Finite product of compact subspaces is compact I am working through an old qualifying exam and I have run into what I think is a very simple question but the way it is posed is confusing to me and I am worried I a missing the point.  The question is:

Assume as known that a finite product of compact spaces is itself compact.  Let $K_i$ be a compact subspace of the space $X_i$, $i=1,...,n$.  Show that $K_1\times...\times K_n$ is a compact subspace of $X_1\times...\times X_n$.

This seems like a silly question.  

It is my understanding that a subspace of a space is compact iff it is a compact space in the subspace topology.  So since each $K_i$ is a compact subspace, each $K_i$ is a compact space in the subspace topology.  But we are to assume that finite products of compact spaces are compact.  Hence, $K_1\times...\times K_n$ is a compact space in the subspace topology, so a compact subspace of $X_1\times...\times X_n$.  

Is there something more subtle going on here that I am missing?
I would assume they want you to prove that the subspace topology induced by the product topology is the same as the product topology of the induced subspace topologies.
Something does seem rather strange in the question as stated.  It might be more convincing to actually prove what the statement is telling you to assume.  It's actually not hard to prove that $K_1 \times \ldots \times K_n$ is compact in $X_1 \times \ldots \times X_n$ provided that each $K_i \subseteq X_i$ is compact.  Let $\mathcal{U} = \{U_i \; | \; i \in I\}$ be some open cover for $K_1 \times \ldots \times K_n$, and note that this induces an open cover for each $K_j$, namely we can write $\mathcal{U}_j = \{U_i \cap X_j \; | \; i \in I\}$.  This induces a finite subcover over $K_j$ by picking out some finite set of indices $I_j \subseteq I$.  We can therefore let $\widetilde{\mathcal{U}}_j = \{U_i \; | \; i \in I_j\}$.  The union

$$
\bigcup_{j=1}^n \widetilde{\mathcal{U}}_j
$$

induces a finite subcover over $K_1 \times \ldots \times K_n$.