The fact that various finiteness conditions lead to good theorems which are manifestly false in their absence seems like a good explanation of why they are important. (In fact, I am having trouble thinking of a wholly different kind of explanation for why anything in pure mathematics is important.)
I think you are on to something to the extent that we need to give nonexamples and counterexamples along with our theorems in order to give students even a fighting chance at appreciating them. In the realm of commutative algebra this was something that was notoriously underappreciated until relatively recently: I recall well Rota writing about the "hygienic theorems" [Rota, Indiscrete Thoughts, pp. 215-216] in algebra, e.g. things like "Every regular domain is normal". As he wrote, we have no chance of grasping results like this unless we see examples -- preferably several -- of domains which are not regular, not normal, and normal but not regular. In this particular example this is easily done, but unfortunately many of the core counterexamples in the subject have a reputation of being too difficult to show beginners. At this point I feel the need to quote directly from p. 136 of Reid's Undergraduate Commutative Algebra:
The catch-phrase "counterexamples due to Akizuki, Nagata, Zariski, etc. are too difficult to treat here" when discussing questions such as Krull dimension and chain conditions for prime ideals, and finiteness of normalisation is a time-honoured tradition in commutative algebra textbooks (comparable to the use of fascist letters $\mathfrak{P}$ and $\mathfrak{m}$ etc., for prime and maximal ideals). This does little to stimulate enthusiasm for the subject, and only discourages the reader in an already obscure literature; I discuss here three counterexamples (taken, with some simplifications, from the famous "unreadable" appendix to [Nagata]) to show some of the ideas involved.
This is very well said (well, except that I honestly don't know what's wrong with $\mathfrak{m}$...): most of the standard texts in commutative algebra leave unanswered the natural questions an alert reader will have: is this hypothesis necessary? is the converse of this result true? What happens if we don't assume that $M$ is a finitely generated module over a Noetherian domain? and so forth.
By a coincidence I have just finished -- that is, within the last half hour -- teaching a first graduate course on commutative algebra. I tried to spend a lot of time on examples, and I was not afraid to make "technical" digressions about what happens when $M$ is not a finitely generated....Especially I spent an extra long amount of time on module-theoretic questions, which made me feel closer to the heart of the subject. It is easy to motivate the need for modules to be finitely generated: there is a structure theorem for finitely generated modules over a PID but there is no structure theorem for infinitely generated abelian groups. The example of $\mathbb{Q}_p$ as a $\mathbb{Z}_p$-module shows that even over a DVR infinitely generated modules can have a complicated structure. Then, when I got to Noetherian rings I motivated them in part by showing that the Noetherian condition was equivalent to many seemingly innocuous and desirable properties, like every submodule of a finitely generated module being finitely generated. At the same time I discussed plenty of examples of non-Noetherian rings, including rings which are very nice "except that they are non-Noetherian" like the ring of all algebraic integers. So I think I gave my students at least an opportunity to feel their way around finiteness conditions in the subject.
Let me add that there are some recent texts which do a much better job at this. Most of all I can enthusiastically recommend T.Y. Lam's Lectures on Modules and Rings. As with all of his books, his skill at balancing theory and examples is superior and makes for very pleasant, stimulating reading.
It goes much the same for compactness in elementary analysis, but it seems easier to me to supply the necessary counterexamples: every time you encounter a theorem which holds on a compact interval $[a,b]$, ask yourself whether it holds on noncompact intervals (and, if applicable, compact non-intervals!). In all the instances I can think of now, such counterexamples are well known and relatively easy to supply.
Extending on Ralph's answer, there is a similar very neat proof for the formula for $Q_n:=1^2+2^2+\dots+n^2$. Write down numbers in an equilateral triangle as follows:
1
2 2
3 3 3
4 4 4 4
Now, clearly the sum of the numbers in the triangle is $Q_n$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$
(I first heard this proof from János Pataki).
How to prove formally that all positions sum to $2n+1$? Easy induction ("moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases"). This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.
How to generalize to sum of cubes? Same trick on a tetrahedron. EDIT: there's some way to generalize it to higher dimensions, but unfortunately it's more complicated than this. See the comments below.
If you wish to tell them something about "what is the fourth dimension (for a mathematician)", this is an excellent start.
Best Answer
This is always helpful if the proof is long and complicated.
Some experts may need only this to understand the proof. Some people may read only some steps, once they have the general idea. Those who prefer to read only the full detail, can always skip this informal description. So I do not see any arguments against this, except the increased length of the paper. But this is rarely a problem. I usually do this in long papers.