The proof of Zorn's Lemma In the book Halmos's proof of Zorn's Lemma. it says that if $C$ is a chain in $\mathbb{X}$(the collection of chain in $X$), then the union of the sets in $C$ belongs to $\mathbb{X}$. I don't understand what is called the sets in $C$
 A: w.r.t. Halmos's proof:
X is a partially ordered set.
$\mathbb{X}$ is the collection of all chains in X. That means that the elements of $\mathbb{X}$ are subsets of X each of which is a chain in X  (i.e. each is totally ordered - all elements are comparable).
In $\mathbb{X}$ itself the elements, which are subsets of X,  can be partially ordered by inclusion, and so you can consider chains in $\mathbb{X}$. If C is such a chain then it's comprised of elements of $\mathbb{X}$, which are in fact subsets of X.
Good luck with the rest of the proof. I found it useful to construct some examples: e.g. start with $X = ${$ 1, 2, 3, a, b$} with ordering $1 < 2 < 3$ and $a < b$. Then $\mathbb{X} $ = { {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, {a}, {b}, {a, b} }. Chains in $\mathbb{X}$ include {{1}, {1, 2}}, {{2}, {2, 3}, {1, 2, 3}}, etc. In these examples, you can see that the union of a chain in $\mathbb{X} $ is indeed a chain in $X$:  $\cup$ {{2}, {2, 3}, {1, 2, 3}} = {1, 2, 3} is a chain in X and therefore an element of $\mathbb{X}$.
A: I think all this is stating is that the union of a chain (w.r.t. inclusion) of chains of elements in the partial order $X$ is again a chain of elements of $X$.
Below is my entire writeup of the proof.
Proof (Zorn's Lemma). Let $\mathscr{X}$ be the
set of all chains in $X$. The collection $\mathscr{X}$ is a nonempty
collection of sets, partially ordered by inclusion. For each chain
$\mathscr{C}$ in $\mathscr{X}$ , its union is in $\mathscr{X}$;
this is just a natural fact, not a result of the chain hypothesis.
But now finding a maximal set in $\mathscr{X}$ leads to a maximal
element of $X$. 
Suppose $Y$ is a maximal set in $\mathscr{X}$. As $Y$ is a chain
in $X$, it has an upper bound $x$. We show that $x\in Y$ (and as
it is an upper bound of Y it follows that x is the maximum element
of Y) and that x is a maximal element of X. Suppose that $z\geq x$
holds for some $z\in X$. Note that $Y\cup\{z\}$ is also a chain
in X, and so by maximality of Y it must be that $Y\cup\{z\}=Y$, and
therefore $z\in Y$. (In particular, as $x\geq x$ we have $x\in Y$.)
As x is an upper bound of Y we have $z\leq x$, and so $z=x$. Thus
x is a maximal element of X.
The family $\mathscr{X}$ of all chains in the partially
ordered set $\mathscr{X}$ satisfies the critical conditions: 
(i) $\mathscr{X}$ is non-empty; $\emptyset\in\mathscr{X}$
(ii) Every subset of an element of $\mathscr{X}$ is also
an element of $\mathscr{X}$. 
(iii) The union of every chain of elements of $\mathscr{X}$
is an element of $\mathscr{X}$. 
For each set $A$ in $\mathscr{X}$, let $\hat{A}$ be the
set of all elements $x$ of $X$ whose adjuction to $A$ produces
a set in $\mathscr{X}$. That is, $\hat{A}:=\{x\in X:A\cup\{x\}\in\mathscr{X}\}$.
Let $f$ be a choice function for $X$. That is, $f$ is a function
from the collection of all nonempty subsets of $X$ to $X$ such that
$f(A)\in A$ for all $A_{\neq\emptyset}\subset X$. Define a function
$g$ from $\mathscr{X}$ to $\mathscr{X}$ by $g(A):=A\cup\{f(\hat{A}\setminus A)\}$
if $\hat{A}\setminus A\neq\emptyset$, $g(A):=A$ otherwise. What
we must prove is that there exists in $\mathscr{X}$ a set $A$ such
that $g(A)=A$. This will be our maximal set in $\mathscr{X}$.
If there is a set in $\mathscr{X}$ to which nothing else
can be added (and remain in $\mathscr{X}$), then it is a maximal
set in $\mathscr{X}$. Why? Well if not a maximal set, then one set
in $\mathscr{X}$ is not contained in it, which means there is an
$x\in X$ that is not in it. If $x$ is comparable to something in
$A$, we could add it and get a bigger chain. If $x$ is not comparable
to anything in $A$, also nothing in $x$'s chain is comparable to
$A$... There could be multiple maximal sets, just are totally incomparable.
we want ultimately a maximal element in $X$. Any maximal set will
lead to one; a maximal element is just something that is not less
than anything else in $X$.
Suppose we have a subcollection $\mathfrak{J}$ of $\mathscr{X}$
that satisfies the $tower$ properties:
(i) $\emptyset\in\mathfrak{J}$
(ii) if $A\in\mathfrak{J}$, then $g(A)\in\mathfrak{J}$
(iii) if $\mathscr{C}$ is a chain in $\mathfrak{J}$, then $\bigcup\mathscr{C}\in\mathfrak{J}$.
Note: As for $\mathscr{X}$, by a chain in $\mathfrak{J}$
we mean ordered by the inclusion relation.
$and$ is a chain. Then by (iii) the union $A$ of all its
sets is itself a set in $\mathfrak{J}$. And by (ii) $g(A)\in\mathfrak{J}$,
so $g(A)\subset A$. Since always $A\subset g(A)$, we have $g(A)=A$,
and we're done!
Note: Towers exist; $\mathscr{X}$ is one. Since the intersection
$\mathfrak{J_{0}}$ of all subcollections of $\mathscr{X}$ with the
tower properties satisfies the tower properties (easily verified),
clearly this will be our best hope for something that is also a chain
(our desired $\mathfrak{J}$). Why? Taking away elements of a chain
always leaves a chain. 
Showing $\mathfrak{J_{0}}$ is a chain is not so simple.
Say that a set $C$ in $\mathfrak{J_{0}}$ is $comparable$
if it is comparable with every set in $\mathfrak{J_{0}}$; this means
that if $A\in\mathfrak{J_{0}}$ then either $A\subset C$ or $C\subset A$.
To say $\mathfrak{J_{0}}$ is a chain means that all the sets in $\mathfrak{J_{0}}$
are comparable. Comparable sets exists; $\emptyset$ is one. 
Let $C$ be a comparable set. Consider the collection $\mathscr{U}$
of all sets $A\in\mathfrak{J_{0}}$ for which either $A\subset C$
or $g(C)\subset A$. The collection $\mathscr{U}$ is somewhat smaller
than the collection of sets in $\mathfrak{J_{0}}$ comparable with
$g(C)$. Anyway, $\mathscr{U}$ it is a tower in $\mathfrak{J_{0}}$.
Fact: If $A\in\mathfrak{J_{0}}$ and $A\subset C$ proper, then $g(A)\subset C$.
To see this just note $g(A)\in\mathfrak{J_{0}}$ and $g(A)$ can have
at most one point more than $A$ (our earlier use of AC in the definition
of $g$ was critical).
Since $\emptyset\subset C$, (i) is satisfied, and (iii) follows by
definition of $\mathscr{U}$. Let us prove (ii) by considering the
cases:
Case 1: $A\subset C$ proper. By the fact above, $g(A)\subset C$
so $g(A)\in\mathscr{U}$.
Case 2: $A=C$. Well, $g(A)=g(C)\in\mathscr{U}.$
Case 3: $g(C)\subset A$. Then $C\subset g(A)$ so $g(A)\in\mathscr{U}$.
Therefore, since $\mathfrak{J_{0}}$ is the smallest tower,
$\mathfrak{\mathscr{U=}J_{0}}$. This shows for any comparable set
$C$, the set $g(C)$ is comparable also. For, given any $C$ we can
form $\mathscr{U}$ as above. The fact that $\mathscr{U}=\mathfrak{J_{0}}$
means that if $A\in\mathfrak{J_{0}}$ then either $A\subset C$ (in
which case $A\subset g(C)$) or $g(C)\subset A$. 
We now know that (i) $\emptyset$ is comparable, (ii) $g$
maps comparable sets to comparable sets, and (iii) the union of a
chain of comparable sets is comparable. Thus the comparable sets in
$\mathfrak{J_{0}}$ constitute a tower and hence exhaust $\mathfrak{J_{0}}$.$\blacksquare$
