"A straightforward application of Zorn's Lemma"? I was reading a textbook on functional analysis and I came across the following:

4.2. Proposition. If $\mathscr{E}$ is an orthonormal set in $\mathscr{H}$, then there is a basis for $\mathscr{H}$ that contains $\mathscr{E}$.
The proof of this proposition is a straightforward application of Zorn's Lemma and is left to the reader.

It is assumed $\mathscr{H}$ is a Hilbert space.
This caught me completely off guard. I've never heard of Zorn's Lemma (though I probably should've...), and when I looked it up it was some bizarre criteria for partially ordered sets to have a maximal element. Furthermore, it's equivalent to AC? You can't just drop a lemma like that and expect everyone to be ok with it.
Am I missing something? Is there another "Zorn's Lemma" that they're referring too? If not, what's the "straightforward application" they're talking about, because I don't see it at all.
By the way, the textbook is Conway's "A Course in Functional Analysis", 2nd Ed.
 A: You have the right Zorn's Lemma. The partial orders we consider when applying Zorn usually consist of subsets of the object we're trying to build, and the maximal elements are the completed objects.
The standard example is bases of a vector space $V$. To make things interesting, let's assume $V$ is "large" - say, infinite dimensional (= no finite basis). How are we going to build a basis for $V$? 
We use Zorn. Here our poset $\mathbb{P}_V$ consists of all linearly independent sets, ordered by "$\subseteq$". If I have a chain of linearly independent sets, their union is linearly independent (this requires that they form a chain under $\subseteq$ - obviously the union of linearly independent sets is not, in general, linearly independent); this means that $\mathbb{P}_V$ satisfies the hypotheses of Zorn's lemma, since the union of a chain is an upper bound of the chain. And this means there is a maximal element of $\mathbb{P}_V$. It's not hard to show that a maximal linearly independent set is a basis - so we've proved that every vector space has a basis. (Note that Zorn says nothing about unique maximal elements, merely that there is at least one maximal element - indeed, usually there will be lots of maximal elements, as in this case.)
Do you see how to use similar reasoning here? (Note: I suspect "basis" should be "orthonormal basis".)
Incidentally, the statement "Every vector space has a basis" is also equivalent to the axiom of choice, and hence to Zorn's lemma; the hard direction was shown by Andreas Blass.

As for motivating Zorn's lemma, you're right, it's often dropped into textbooks out of thin air. And it is very weird; there's a standard joke that "The axiom of choice is obviously true, the well-ordering principle is obviously false, and who can tell about Zorn's lemma" (in fact the three statements are equivalent). The proof of Zorn's lemma is definitely nontrivial; see here. So often it is used as a blackbox, because it is extremely useful.
A: Consider $L$ the set of orthonomal set which contains the orthonormal set $E$ ordered by inclusion.  Every family $(E_i)_{i\in I}$ of totally ordered subsets of $L$ has a sup element $\bigcup_{i\in I}E_i$, so you have a maximal element$L_0$ of $L$ which is a basis. Suppose that $L_0$ is not a basis and consider the vector subset $V$ generated by $L_0$, you have $x\in H$ not in $V$. Let $W$ the orthogonal subset of $V$, you can write $x=x_1+x, x_1\in V, x_2\in W$, $x_2\neq 0$, you have $L_0\bigcup \{x_2\}$ contains strictly $L_0$ contradiction since $L_0$ is maximal.
