Why we can't say $]1,2]$ has a Minimum? Is it because we use this assertion only in the context of a partial order or total order which are not strict?
The Well-Ordering-Theorem says that every set can be well-ordered. A set X is well-ordered by a strict total order if every non-empty subset of X has a least element under the ordering.
And the Concept of Minimum is based on boundedness. A subset of a partially ordered set X is called bounded if it has both an upper and a lower bound.
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It does not have a minimum because the infimum of the set, which is $1$, is not contained in $\left]1,2\right]$.
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You are mixing the well-ordering theorem where you shouldn't. That theorem is applicable if you already have a set, but you don't have the ordering on it yet. The theorem then claims that a well ordering on that set exists.
On $(1,2]$ we cannot use that theorem because we are already given ordering $\le$, which is the usual ordering of real numbers. That ordering is not a well ordering, and there are certain subsets (such as all of $(1,2]$) that don't have minimum. There is, certainly, another ordering on $(1,2]$, which will be well ordering ... but it's not $\le$, it is something else.
If we are going to discuss the minimum of $(1,2]$ or sets like it in this context, we would do well the specify with respect to which ordering. In the case of the usual total ordering $\le $ on $\mathbf{R}$, it's fairly easy to see that no minimal element in $(1,2]$ can exist. Suppose that $1+\varepsilon$ were a minimal element in $(1,2]$. Then $1+\frac{\varepsilon}{2}\in (1,2]$ and $1+\frac{\varepsilon}{2}<1+\varepsilon$, contradicting minimality of $1+\varepsilon$.
It's worth noting that the Well-Ordering Theorem only provides the existence of an ordering with respect to which a given set $X$ is well-ordered. This may not coincide with a pre-existing order that the set $X$ may have been equipped with.