TITLE: understanding schemes finite over Spec $K$
QUESTION [9 upvotes]: I am following Vakil's FOAG, exercise 7.3.H:
Let $X\to $Spec $K$ be a finite morphism, prove that $X$ is a finite union of points with the discrete topology. 
I am following the guidance there. If we write $X=$Spec $A$ then $A$ is a finite dimensional vector space over $K$. If $A$ was a domain then it is easy to show that it is a field, and so we get that all primes of $A$ are maximal, and hence $X$ consists only of closed points. 
The next part should be to prove that $X$ is discrete, and then finiteness would follow from quasicompactness.
My question is why is $X$ discrete? I will be glad for anything you can say about the general problem, but I am looking to understand how it is possible to show discreteness now, before finiteness, say.

REPLY [9 votes]: (An answer based on an exchange of comments with the OP.)
As the OP observes, a finite-dimensional $K$-algebra that is a domain is necessarily a field, and so all prime ideals in $A$ are maximal.
Now, by CRT, if $\mathfrak m_1, \ldots,\mathfrak m_k$ are distinct maximal ideals,
then $$A/(\mathfrak m_1 \cap \cdots \cap \mathfrak m_k) \cong A/\mathfrak m_1 \times \cdots \times A/\mathfrak m_k, $$ and so $k \leq \dim_K A.$  In particular,
$A$ admits no more than $\dim_K A$ maximal ideals, and so Spec $A$ is a finite
set of closed points.  

Unfortunately, I don't see how to directly follow the hint (i.e. to first prove discreteness) in a natural way.