You are thinking about finite subsets. And it is true, given a finite subset of a linearly ordered set, it is has a minimal (and maximal) element.
But what about infinite subsets? What about $\Bbb Q$, for example, as a subset of $\Bbb R$ or as a subset of itself?
And even more so, your argument if you look closely, should work for maximal. Every two elements are comparable, then there is a maximal element to every non-empty subset. But surely you can find linear orders without a maximal element, even well-orders without a maximal element, e.g. $\Bbb N$.
A well-ordering, as you say, is a linear ordering where every nonempty set has a least element. Every well-ordering is a linear ordering by definition$^*$, but the converse is not true - the following are examples of linear orders which are not well-ordered:
$\mathbb{Z}$ ($\mathbb{Z}$ itself has no least element).
$\{{1\over n+1}: n\in\mathbb{N}\}\cup\{0\}$ (while the whole set does have a least element, the nonempty proper subset $\{{1\over n+1}: n\in\mathbb{N}\}$ does not).
$\mathbb{Q}$, $\mathbb{R}$, $[0, 1]$, ... There are lots.
The simplest examples of well-orderings are the finite linear orders and $\mathbb{N}$ itself (the fact that $\mathbb{N}$ is well-ordered is the thing that makes proof by induction work!). However, there are bigger well-orderings; for example, "$\mathbb{N}+\mathbb{N}$," where "$+$" denotes the sum of linear orders (put the first "after" the second). Concretely, an example of a linear ordering of type $\mathbb{N}+\mathbb{N}$ would be $$\{1-{1\over n+1}: n\in\mathbb{N}\}\cup\{2-{1\over n+1}: n\in\mathbb{N}\}$$ with the usual ordering.
Indeed, there are arbitrarily large well-orderings, even though they get increasingly difficult to visualize. For any infinite cardinal $\kappa$, the set of isomorphism types of well-orderings$^{**}$ of cardinality $\le \kappa$ is itself well-ordered by "embeds into," and has cardinality $>\kappa$ (in fact, $\kappa^+$). For a concrete example, there are uncountable well-ordered sets! Note that this does not rely on the axiom of choice.
$^*$Actually, we can say a bit more: any antisymmetric relation $R$ on a set $X$ satisfying $$\mbox{For all nonempty $Y\subseteq X$, there is some $y\in Y$ with $yRz$ for all other $z\in Y$}$$ is actually a well-ordering (that is, the "linear ordering" requirement is superfluous): we already have antisymmetry, so now just show trichotomy and transitivity:
For trichotomy, given $x\not=y$ think about the two-element set $\{x, y\}$ ...
For transitivity, suppose $xRy$ and $yRz$ (note: by antisymmetry this means $x\not=z$) but $x\not Rz$. By trichotomy, we have $zRx$. But now think about the three-element set $\{x, y, z\}$ ...
$^{**}$There's a bit of an issue here actually: an isomorphism type is a proper class, so we can't form the set of isomorphism types of well-orderings of a given cardinality. There are various ways to get around this, and it's at this point that the von Neumann ordinals should be introduced. But this should really be a side issue.
Best Answer
If $\leq$ is a total order on a set $S$, then the new relation $<$ defined by $x < y$ iff ($x \leq y$ and $x \neq y$) is a strict total order on $S$.
If $<$ is a strict total order on a set $S$, then the new relation $\leq$ defined by $x \leq y$ iff ($x < y$ or $x = y$) is a total order on $S$.
In other words, in a fairly evident way one can always exchange a total order for a strict total order and conversely. So it doesn't really matter which definition is taken. One can easily check that the definition of a well-order in one setting carries over to the definition of a well-order in the other setting.