Let $E$ be totally ordered under $<$ , and dense. I would like to prove that a well ordered set is never dense.
To prove this, I tried to fid a non-empty subset $X$ of $E$ such that there is no least element. So, I take $X=\{z\in E: \forall x,y\in E\quad x<z<y\}$, it's non-empty because E is dense.
If I assume that there is a least element, noted $l$. We have $$\forall w\in X\quad l<w.$$ As $l\in X$ and $<$ is dense, we can find an element between $x$ and $l$, therefore $l$ is not a least element.
Is it correct?
As fleablood commented, this isn't quite right because your definition of $X$ isn't what you want it to be. For $z$ to be an element of $X$, $z$ has to satisfy $x<z<y$ for every $x,y\in E$ simultaneously, which is impossible! For instance, you might take $x=y=z$, in which case $x<z<y$ is definitely false.
What you want to do instead is fix two elements $x,y\in E$ with $x<y$ ahead of time and then define $X$ using these two fixed elements, without a quantifier. Note that this raises an important point: the statement you are trying to prove is not actually true in general, since you need to know that two such elements $x$ and $y$ exist! If $E$ is empty or has only one element, then in fact $E$ is both well-ordered and dense.