Show that the ordered set $(0,1)$ is a substructure of $(0,1]$ in the language $L=\{<\}$ but is not an elementary substructure of it in the same language.
Since the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$ , it's also a substructure of it.
Now, $\varphi: (\exists x)(x=1)$ is satisfied in $(0,1]$ but not in $(0,1)$ so, $(0,1)$ is not an elementary substructure of $(0,1]$.
So, I wonder if saying "the ordered set $(0,1)$ is a subset of the ordered set $(0,1]$" is enough to prove that $(0,1)$ is a substructure of $(0,1]$.
As spaceisdarkgreen points out, what you've done isn't correct: the language $L$ of pure linear order doesn't have a symbol for $1$, so "$\exists x(x=1)$" is not a sentence in the relevant language.
What you'll need to do is somehow define $1$ in the language of linear order alone; do you see a special property which $1$ has, in the structure $(0,1]$?
It's worth pointing out a subtlety of the notion of substructure: as long as our language has only relation symbols, then any subset of our structure is (or rather, is the underlying set of) a substructure, but this is no longer true if our language includes function symbols - e.g. $\{$odd numbers$\}$ is not (the underlying set of) a substructure of $(\mathbb{N}; +)$, since it's not closed under $+$. In fact, even constant symbols (= zero-ary functions) kill this, since a subset not containing the thing named by a given constant symbol won't be a substructure.
It's also worth noting that you've in fact shown that $(0,1)$ is not even elementarily equivalent to $(0,1]$, and that this is actually a stronger property than not being an elementary substructure: e.g. $(0,{1\over 2}]$ is elementarily equivalent (in fact, isomorphic) to $(0,1]$, but is not an elementary substructure of it (why not?).