Let $(A,<)$ an ordered set which means that "$<$" defines a strict order. We consider $B\subseteq A$ a non empty-set and we have to prove that $B$ has at most one minimum.
Here is my attempt. By absurd let's consider $n$ and $n'$ two minimums of $B$. That means that for all $x\in B$ with $x\ne n$, $x>n$. So we can take $x=n'$ and it gives that $n'>n$ but $n'$ is a minimum of $B$ that's why it's impossible to have $n'>n$. But I don't know if this proof works.
Then I have to prove that if $(A,<)$ is a totally ordered set, all minimal elements are the minimum.
We just know that if the set is totally ordered we can compare every elements and in particular minimal elements.
Thanks in advance !
A minimum element for $B$ is $m\in B$ such that, for all $x\in B$, if $x\ne m$, then $m<x$.
Now suppose $m\in B$ and $n\in B$ both satisfy the condition. If $m\ne n$, by definition of minimum we have $m<n$ ($m$ is a minimum) and $n<m$ ($n$ is a minimum). Thus, by transitivity, $m<m$, which is a contradiction to $<$ being a strict order.
If $<$ is a total order and $m$ is a minimal element of $B$, then $m$ is the minimum. Indeed, if $x\in B$ and $x\ne m$, by minimality we know that $x\not<m$; since the order is total, it follows $m<x$ (because one of $x<m$, $x=m$ or $m<x$ must hold).