A paracompact space is a space in which every open cover has a locally finite refinement.
A compact space is a space in which every open cover has a finite subcover.
Why must the product of a compact and a paracompact space be paracompact?
I really have very little intuition about how to go about this question, so any hints or a proof would be greatly appreciated.
On
You could prove that the product of a paracompact space and a compact space is paracompact by using the tube lemma.
I think Trevor is right. Here are the details (this is an adaptation of the classical proof that the product of two compact spaces is a compact space that you can find in Munkres, for instance).
Let $X$ be a paracompact space, $Y$ a compact one and ${\cal U}$ an open cover of $X \times Y$.
For any $x \in X$, the slice $\left\{ x \right\} \times Y$ is a compact space and ${\cal U}$ an open cover of it (as a subspace). So it admits a finite subcover $U_{x,1}, \dots , U_{x,n_x} \in {\cal U}$. Let us call $N_x = U_{x,1} \cup \dots \cup U_{x,n_x}$ their union.
So $N_x$ is an open set that contains the slice $\left\{ x \right\} \times Y$. Because of the tube lemma, there exists an open set $W_x \subset X$ such that
$$ \left\{ x \right\} \times Y \quad \subset \quad W_x \times Y \quad \subset \quad N_x \ . $$
Now, those $W_x$ form an open cover ${\cal W}$ of $X$. By hypothesis, there is a locally finite refinement ${\cal W}' = \left\{ W_{x_i}\right\}$, $i\in I$ for some index set $I$.
Consider the following subcover of ${\cal U}$:
$$ {\cal U}' = \left\{ U_{x_i,j}\right\} \ , $$
with $i\in I$ and $j = 1, \dots , n_{x_i}$.
Let us show that ${\cal U}'$ is a locally finite subcover of ${\cal U}$: take any point $(x,y) \in X \times Y$. By hypothesis, there is a neighborhood $V \subset X$ of $x$ such that $V$ interesects only finitely many of the sets of ${\cal W}'$. Then $V\times Y$ is a neighborhood of $(x,y)$ that intersects only finitely many of the sets of ${\cal U}'$.