I am having some trouble with the following problem from Terence Tao's Analysis 1 (Third Edition)
Let $X$ be a set, and let $\Omega$ be the space of all pairs $(Y,\leq)$, where $Y$ is a subset of $X$ and $\leq$ is a well-ordering of $Y$. If $(Y,\leq)$ and $(Y',\leq ')$ are elements of $\Omega$, we say that $(Y,\leq)$ is an initial segment of $(Y',\leq ')$ if there exists an $x\in Y'$ such that $Y=\{y\in Y': y<'x\}$ (so in particular $Y\subset Y'$), and for any $y,y'\in Y$, $y\leq y'$ iff $y\leq ' y'$. Define a relation $\preceq$ on $\Omega$ by defining $(Y,\leq)\preceq (Y',\leq ')$ if either $(Y,\leq)=(Y',\leq ')$, or if $(Y,\leq)$ is an initial segment of $(Y',\leq ')$.
I've shown that $\preceq$ is a partial ordering on $\Omega$. Now I am trying to use Zorn's lemma to show that there is a maximal element which will then show that the Well-Ordering Theorem is true. This is what I have so far
Let $Z$ be a totally ordered subset of $\Omega$. We define the set $Y=\bigcup_{(S,\leq_S)\in Z}S$ and a relation on Y $\leq_Y$ defined as follows: $y\leq_Yy'$ if and only if there exists a $(S,\leq_s) \in Z$ such that $y,y' \in S$ and $y \leq_s y'$. It is not hard to verify that this is a partial ordering on $Y$ and in fact it is a total ordering on $Y$.
I am having hard time figuring out how to show that this is a well-ordering on $Y$. I have tried to show that if $W$ is a non-empty set of $Y$ then there is a $(S,\leq_S)\in Z$ such that for all $w \in W$ we have $w \in S$ so in particular there is a $w \in W$ such that $w \leq_s w'$ for all $w' \in W$ and this would imply that $W$ has a minimal element. However I am struggling to make this idea work as well.
I will be very grateful for any help that is offered. In particular how could I go about showing that $\leq_Y$ is a well-ordering on $Y$ (if it is at all)?
If $W \subseteq Y$ is non-empty, pick $w_0 \in W$ and $(S_0,\le_{S_0}) \in Z$ so that $w_0 \in S_0$. Then $W \cap S_0$ is a non-empty subset of the well-order $(S_0,\le_{S_0})$ so has a minimum $m \in S_0 \cap W$ for $\le_{S_0}$.
Now, if $w \in W$ is arbitrary, we know it's in some $(S,\le_S) \in Z$ by definition, and we have (as $Z$ is linearly ordered under $\le$), either $(S_0,\le_{S_0}) \le (S,\le_S)$ or the other way around. If $(S, \le_S) \le (S_0,\le_{S_0})$ we have that $w \in W \cap S \subseteq W \cap S_0$ so that $m \le_{S_0} w$ and thus $m \le_Y w$ follows immediately. If $(S_0,\le_{S_0}) \le (S,\le_S)$ holds, we have $m \in S$ and $m \le_S w$ because $w <_S m$ would have meant that $w \in S_0$ too (because we initial segments !) and that would have contradicted minimality in $S_0$.
So $(Y, \le_Y)$ is a well-order. Now show that $(Y, \le_Y)$ is not a proper initial segment of any $(S, \le_S) \in Z$, to see it is a maximal element of $Y$ in $\Omega$.