Given x in X, the "weak initial segment of x", ws(x), is

ws(x) = {y in X : y<=x}.

You can consider ws as a function from X to P(X), the power set of X;
the power set of X is ordered by inclusion, and we have that ws(x) is
contained in ws(y) if and only if x<=y, so ws can be viewed as an
order-preserving function between the partially ordered set X and the
partially ordered set P(X). Morevoer, the function ws is one-to-one:
if ws(x) = ws(y), then since x is in ws(x) and y is in ws(y), it
follows that x<=y and y<=x, so x=y.

So, Halmost says: "finding a maximal element in X is the same as
finding a maximal element in" the range of ws. Because you have an
order isomorphism between X and its image under ws. And since X
satisfies that every chain in X has an upper bound in X, then every
chain in the range of ws has an upper bound in the range of ws.

Now let Y be the set of all chains in X, partially ordered by
inclusion. Note that if A is a chain, then A has an upper bound x in
X, so therefore, A is contained in ws(x) [here we are using the
hypothesis on X]. Thus, every element of Y is dominated by an element
of the range of ws.

This means that if y is a maximal element in Y (with respect to the
inclusion order of Y), then y has to be a maximal element in ws(x) as
well. For if y is contained in ws(x), then y\/{x} would be a chain,
hence an element of Y; since it contains the maximal element y of Y,
we must have y = y\/{x}, so x lies in y. Thus, y is in fact in the
range of ws, and is also maximal in the range of ws. Thus, finding an
maximal element in Y gives a maximal element of the range of ws, which
in turn gives a maximal element of X.