Please help me locate errors :(
Proposition(Zorn's Lemma): Let $X\neq\emptyset$ be of partial order with the property that $\forall Y\subseteq X$ such that $Y$ is of total-order then $Y$ has an upperbound, then $X$ contains a maximal element.
Proof: Case 1: $B\neq\emptyset$ such that $B$=$\{$$b\in X$: $b$ has an undefined order relation with all elements in $X$$\}$. Thus, $\forall b\in B$, $b$ is maximal since there is no $x\in X$ where $b<x$.
Case 2: $B=\emptyset$; Let $I_1$ be some index set and let $Y_\alpha,\forall\alpha\in X_1$ denote all the totally-ordered sets in $X$. Consider that $\cup_{\alpha\in I_1}Y_\alpha$ is either a poset or a totally-ordered set in $X$. If $\cup_{\alpha\in I_1}Y_\alpha$ is totally ordered, then, by the premise of the lemma, $\cup_{\alpha\in I_1}Y_\alpha$ has an upperbound $\hat{y}$. Since $\hat{y}$ would then be an upperbound of all totally-ordered subsets of $X$, then no $x\in X$ satisfies $\hat{y}<x$. Hence, $\hat{y}$ is a maximal element in $X$. If $\cup_{\alpha\in I_1}Y_\alpha$ is a poset, then $\exists y_1,y_2\in\cup_{\alpha\in I_1}Y_\alpha$ such that the order relation $y_1$ and $y_2$ is undefined. In order to resolve this issue(of being unable to "compare"), we construct a set $\mathscr{T}$ of totally-ordered sets $\Bbb{T}(y)$ where $\forall\overline{y}\in\cup_{\alpha\in I_1}Y_\alpha$, we let \begin{align}\Bbb{T}(\overline{y})=\{y\in\cup_{\alpha\in I_1}Y_\alpha:y\leq_X\overline{y}\quad \lor\quad \overline{y}\leq_X y\}\end{align} Note that the correspondence of each $\overline{y}\in\cup_{\alpha\in I_1}Y_\alpha$ to a set $\Bbb{T}(\overline{y})$ is not one-to-one. Nevertheless, we have a collection $\mathscr{T}=\{\Bbb{T}(y):y\in\cup_{\alpha\in I_1}Y_\alpha\}=\{\Bbb{T}(y):y\in X\}$. Each $\Bbb{T}(y)$ is constructed in such a way so that $\forall\alpha\in I_1, \exists\Bbb{T}(y)\in\mathscr{T}$ such that $Y_\alpha\subseteq\Bbb{T}(y)$. Moreover, since $B=\emptyset$ and $\forall y\in\cup_{\alpha\in I_1}Y_\alpha$, $\Bbb{T}(y)$ contains the maximum number of elements from $X$ while remaining totally-ordered, then $\forall x_i\in X$ where $x_i\notin\Bbb{T}(y)$, it must be that $x_i\in\Bbb{T}(y')$ where $y$ and $y'$ have an undefined order-relation. Lastly, since we know that $\forall y\in X, \Bbb{T}(y)\subseteq X$ and is totally-ordered, then, by our premise, $\Bbb{T}(y)$ has an upperbound $\tau$ which I think is some maximal of $X$ because all upperbounds of each distinct $\Bbb{T}(y^*)\in\mathscr{T}$ are pair-wise undefined order relation(or incomparable). So for all upperbound $\tau$ in some $\Bbb{T}(y)$ there does not exist $x\in X$ such that $\tau <x$.
Case1. Assume $X$ is itself totally ordered.
You write: thus for a fixed element $b\in B$, $b$ is maximal.
Well, of course, you assumed $B=\emptyset$. What ever you claim for an element of $B$ is vacuously true, since there are no elements to check the condition against. No further justification is required.
We are still short a maximal element, however.
Case2. Do point out that $X$ does contain totally ordered subsets. For instance, since $X\neq\emptyset$, every singleton subset is totally ordered.
If we observe the union of all totally ordered subsets: $\bigcup Y_j \subseteq X$ and assume this is totally ordered, then we have assumed $X$ is totally ordered. Back to Case1.
The set $\mathbb T(y)$ is a chain in $X$ generated by $y$. $\mathbb T(x) = \mathbb T(y)$ iff $x,y$ are comparable i.e they are situated in the same chain. By premise of ZL the set $\mathbb T(y), y\in X$ is bounded. No maximal element in sight.
Your assumptions are too weak.
An alternative, is to assume the Axiom of Choice and proceed by this proof : AC implies ZL. Study proof 2, for it is far less involved.
Reading your post, I also get the feeling you mis-understand something about ZL. It doesn't say how many maximal elements there are. It certainly doesn't tell you the upper bound itself is a maximal element!