This should follow from the fact that equality is an equivalence relation under any set.
$x=x$, for all $x$
$x=y \to y=x$, for all $x$ and $y$
if $x=y$ and $y=z$, then $x=z$, for any $x$, $y$, and $z$
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
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.