Can Zorn's Lemma be 'inverted' like this:? Let $R$ be a (commutative) ring not equal to $0$. I want to show that the set of prime ideals of $R$ has a minimal element w.r.t. inclusion. 
This may be a wholeheartedly wrong attempt, but I thought I'd try this:
Let $P$ be the set of all prime ideals of $R$. We have chains of prime ideals with respect to inclusion, i.e.
$\mathfrak{p}_1\supset \mathfrak{p}_2\supset\ldots\supset \mathfrak{p}_n\supset\ldots $ 
Now consider the chain $R/\mathfrak{p}_1\subset R/\mathfrak{p}_2\subset\ldots\subset R$. 
This has a maximal element, namely $R$, so if we let $U$ be the set $\{R/\mathfrak{p}_i : \mathfrak{p}_i\: \text{is prime}\}$, then every chain has a maximal element, so if we take the order-reversing chains, there must be a minimal element for each chain, hence $P$ has a minimal element w.r.t. inclusion.
I suspect this may be pretty dodgy logic, but I think I don't have a full understanding of when best to apply the lemma.
 A: There are three questions here.
The first question is whether your argument works. The answer is, not yet: see Zev Chonoles' comment.

Second, whether you can apply Zorn's lemma "upside-down." The answer is yes: if $P$ is a poset, then its "inversion" (just reverse the ordering) $P^*$ is a poset, and if $P^*$ satisfies the hypotheses of Zorn then $P^*$ has a maximal element, hence $P$ has a minimal element. 
Unwinding this, we get: if every descending chain in $P$ has a lower bound, then $P$ has a minimal element.

So third question is, what's the gap? Well, the object you've defined isn't really the "inverse" of the poset of prime ideals - I think you're trying to hard. You can just order the prime ideals by reverse inclusion, which I think is what is going on intuitively with your definition. Then everything works the way it should.
A: Some remarks: You consider only a specific chain that happens to end in $R$. Admittedly, this shows that $R$ is an upper bound for any chain. But doesn't that make the application of Zorn useless - after all you start with a very maximal element! On the other hand, it may happen that $R=R/0$ is not even a valid maximum because it is not the quotient by a prime ideal (if $R$ has zero-divisors). 
That being said, no one prevents you from defining the order on the set of prime ideals as $\mathfrak p\le\mathfrak q:\iff \mathfrak q\subseteq \mathfrak p$. This way a "maximal" prime ideal is one that is inclusion-minimal. However, you still have to show that Zorn applies. That is: Given a chain of prime ideals (a set $\{\,\mathfrak p_i\mid i\in I\,\}$ such that for $i, j\in I$ we have $\mathfrak p_i\le \mathfrak p_j$ or $\mathfrak p_j\le \mathfrak p_i$), does there exist an "upper" bound for the chain (i.e., is there a prime ideal $\mathfrak p$ with $\mathfrak p_i\le \mathfrak p$ for all $i\in I$)? Under the partial order ${\le}:={\supseteq}$, something like $\bigcap _{i\in I}\mathfrak p_i$ would seem to be a natural candidate for an upper bound - but is it?
