Every Non empty subset has a least element implies linear order

Question

Suppose $(A,R)$ be structure where R is a binary relation on $A$. Suppose $A$ has the property that every Non empty subset of $A$ has a least element w.r.t. the relation $R$. Then $R$ is a linear order on $A$.

Proof:

Trichotomy: For any two $x,y \in A$, and $x\neq y$ , consider the two elements subset $\{x,y\}$ of $A$.

Anti-Symmetry: Given $x,y \in A$, $x\neq y$ and $xRy$. Now $yRx$ would imply non existence of least element in $\{x,y\}$.

Transitive: Given $xRy$ and $yRz$.

For $x=y$, it is trivial that $xRz$.

For $x\neq y$:

$x\neq z$, otherwise $yRx$, which contradicts Anti-Symmetry.

Hence by trichotomy, $zRx$ or $xRz$.

For $zRx$ consider the 3-element subset $\{x,y,z\}$, which will not have a least element. Hence $xRz$.

Thus $R$ is a linear order. $\blacksquare$

What is wrong with my reasoning?

This question is inspired by Noah's answer to following question:

https://math.stackexchange.com/questions/2703756/the-difference-between-well-order-and-total-order#:~:text=A%20well%2Dordering%2C%20as%20you,itself%20has%20no%20least%20element).

P.S.: I don't have enough reputation to comment on Noah's answer and ask for clarification.