In general, if $S$ is an equivalence relation on a set $X$, then, the equivalence class of $x\in X$ is defined as:
$$[x]=\{y\in X\mid xSy\}.$$
So, $[x]$ must contain, by its definition, all $y\in X$ such that $ySx$, which means that no subset of $[x]$ can be a class on its own, apart from $[x]$ itself.
Let in our case $A=\{1,12,23,\dots,89\}$ and $B=\{23,23,\dots,89\}$. It is obvious that $A=[1]$ and that $B\underset{\neq}{\subset}A$, so $B$ cannot be an equivalence class.
In general, if $x,y\in X$, then it is true that either:
$$[x]=[y]$$
or
$$[x]\cap[y]=\varnothing.$$
from which we can also imply that $B$ is not an equivalance class.
To prove the former, we think as follows:
For every $x,y$ exactly one of the following is true:
$$xSy\text{ or }x\not Sy$$
where, by $\not S$ we mean that $(x,y)\not\in S$.
If $xSy$ then, for every $z\in[y]$ we have that:
$$zSy\overset{xSy}{\Rightarrow}xSz\Rightarrow z\in[x]$$
so $[y]\subseteq[x]$. Also, for every $z\in[x]$ we have that:
$$xSz\overset{xSy}{\Rightarrow}zSy\Rightarrow z\in[y]$$
so $[x]\subseteq[x]$. So, from the above $[x]=[y]$.
If $x\not Sy$, then, if there exists $z\in X$ such that $x\in[x]\cap[y]$ we get that $z\in[x]$ and $z\in[y]$ so $zSx$ and $zSy$ which implies that $xSy$, which is a contradiction. So, $[x]\cap[y]=\varnothing$.
Using this result, we can see that, if $B$ was an equivalence class, then we should have $B\cap A=\varnothing$ or $B=A$ which is not true, so $B$ cannot be an equivalence class.
Last note!
When we write:
$$A=\{x\in X\mid \text{some property for }x\}$$
where $X$ is some set, we mean that $A$ includes all the elements $x$ of $X$ that do have the named property. So, the equivalence class of $x$ contains all the elements that are equivalent with $x$!
Edit: For more information on the specific equivalance relation we are using here and in similar constructions, see this article.
