Actually, without the reflexivity condition, the empty relation would count as an equivalence relation, which is non-ideal.
Your argument used the hypothesis that for each $a$, there exists $b$ such that $aRb$ holds. If this is true, then symmetry and transitivity imply reflexivity, but this is not true in general.
The wording of the exercise could easily have been made more precise.
I suppose that $L$ is a set of strings in $\Sigma,$ that is, $L \subseteq \Sigma,$
and that when we say $L$ "induces an equivalence relation ... over the set of strings," we really mean an equivalence relation over $\Sigma.$
If $L$ is a set of strings and $x$ is a string, then the definition from the lecture
says that $L - x$ also is a set of strings,
specifically all the strings that you can append to $x$ in order to get a string in $L.$
You could think of it this way: a string $y$ is a "good" suffix of $x$
if you can append $y$ to $x$ and get a string in $L,$
that is, $xy\in L.$
Then $L - x$ as the set of "good" suffixes of $x.$
Note that if $x$ itself is in $L,$ then one of the "good" suffixes is the empty string.
The relationship according to your notes is that $x \equiv_L y$
iff the "good" suffixes of $x$ are exactly the same as the "good" suffixes of $y.$
If $L = \{0, 01, 0101\},$ as in your example, then
$L - 010 = \{1\},$
$L - 01 = \{e, 01\}$ (where $e$ is the empty string),
and $L - 0 = \{e, 1, 101\}.$
But $L - 1 = \emptyset$; if you start a string with $1,$ there is no way to finish the string in order to make it something in $L.$
On the other hand, $L - e = L.$ (Actually that last fact would be true no matter what you chose for the members of $L.$)
So $01 \not\equiv_L 010$ in your example, because $\{e, 01\} \neq \{1\}.$
Likewise $0 \not\equiv_L 01.$
But $1 \equiv_L 11,$ because $L - 1 = L - 11 = \emptyset.$
In fact if you choose any $x$ and $y$ that are not prefixes of any string in $L$
then $L - x = L - y = \emptyset$ and therefore $x \equiv_L y.$
As for proving the equivalence relation, it's a relatively mechanical task.
You need to show that if $x, y, z \in \Sigma,$
then $x\equiv_L y$ implies $y\equiv_L x$ (symmetry),
$x\equiv_L x$ (reflexivity), and
if $x\equiv_L y$ and $y\equiv_L z$ then $x\equiv_L z$ (transitivity).
For example, for symmetry, if the set of "good" suffixes of $x$ is the same as the set of "good" suffixes of $y,$ is it always true that the set of "good" suffixes of $y$ is the same as the set of "good" suffixes of $x$?
The trick is to write this formally, using the symbology from your notes.
Best Answer
Unfortunately, none of the properties of reflexivity, symmetry (the usual term for what you are calling "commutativity"), nor transitivity is inherited by subsets.
Moreover, note that while symmetry and transitivity are not contextual (they depend only on your set of pairs, that is, on your relation), reflexivity is contextual: you need to say what your underlying set is in order to discuss reflexivity: the exact same set of pairs may be reflexive when considered as a relation on a set $A$, but not when considered as a relation on a different set $B$, even if $C\subseteq (A\times A)\cap(B\times B)$.
That is, it is false that if $D\subseteq C$ and $C$ is reflexive (on something), then $D$ is reflexive (on something else). It is false that if $D\subseteq C$ and $C$ is symmetric, then $D$ is symmetric. And it is false that if $D\subseteq C$ and $C$ is transitive, then $D$ is transitive.
For instance, take $A=\{1,2,3\}$, $C=\{(1,1), (1,2), (2,1), (2,2), (3,3)\}$, $D=\{(1,1), (1,2),\}$. Then $C$ is reflexive, symmetric, and transitive, $D\subseteq C$, but $D$ is not reflexive (not on $A$ and not on $\{1,2\} $) and not symmetric. You can probably come up with examples where it is not transitive either.
Of course, I didn't construct $D$ by restriction; but the point is that nowhere did you use the fact that you were constructing your relation by restriction. You just asserted that $C\cap(A_0\times A_0)$ would have the desired properties by virtue of being contained in a relation that did, and that argument is invalid.
What you need to answer:
Reflexivity on $A_0$. Let $a\in A_0$; why must $(a,a)$ be in $C\cap (A_0\times A_0)$ (Hint: Must it be in each of the two sets you are intersecting?)
Symmetry. If $(a,b)\in C\cap(A_0\times A_0)$, why must $(b,a)$ also be in $C\cap (A_0\times A_0)$?
Transitivity. If $(a,b)\in C\cap (A_0\times A_0)$, and $(b,c)\in C\cap(A_0\times A_0)$, why must $(a,c)$ be in $C\cap(A_0\times A_0)$?