algebraic version of "finite covering of a compact space is compact" The following statement is an exercise in point set topology: If $E \to X$ is a covering with nonempty finite fibers and $X$ is compact, then also $E$ is compact. Now Grothendieck generalized covering theory so that in particular separable field extensions may be regarded as coverings.
Question: What is the corresponding statement in field theory?
 A: To answer Martin's questions in his comments to Rayleigh's answer:
Fix a scheme $S$ and a proper morphism $f:X\to Y$ of $S$-schemes. Suppose that $Y$ is proper over $S$. If $f$ is proper, then $X$ is proper over $S$. This is simply because proper morphisms are stable under composition.
How does one prove that proper morphisms are stable under composition? One simply proves this property for finite type morphisms of schemes and separated morphisms. Then you're done.
If you stick to fields, all morphisms are separated so to prove that proper morphisms are stable under composition in this case, you simply have to prove that if you have a tower $K\subset L\subset M$ of finite degree field extensions, then $K\subset M$ is of finite degree. This is an easy fact. In conclusion, the proof of the statement for fields is easy and the statement itself doesn't give any nontrivial information in the case of fields.
I think Rayleigh added the additional hypotheses of "finite etale" to his statement, because he wanted to mimic the set-up of a "covering".
A: Fix a scheme $S$.
Let $f:X\longrightarrow Y$ be a finite etale morphism of $S$-schemes, with $Y$ an integral $S$-scheme.
Note that $f$ is proper.
Therefore, if $Y$ is proper over $S$, we have that $X$ is also proper over $S$.
This is (I believe) the analogue of the statement in your question. (Take $S$ to be the spectrum of a field.)
