Languages with infinitely many but finite Myhill-Nerode equivalence classes
A random language would do. If $L$ is chosen by putting every word in $\Sigma^*$ in $L$ with probability $1/2$, then for every two given words $x \neq y$ and any third word $z$, the probability that $xz \in L \Leftrightarrow yz \in L$ is $1/2$. Hence almost surely $x$ and $y$ are not equivalent (since we can choose an infinite sequence $z_i$ so that the words $xz_1,yz_1,xz_2,yz_2,\ldots$ are all distinct). Since there are only countably many pairs of words, we find that almost surely, every word is its own equivalence class.

It is also not too difficult to explicitly construct such a language. It suffices to find a subset $A$ of the natural numbers such that for every $x,y \in \mathbb{N}$ there exists $z \in \mathbb{N}$ such that $x+z \in A \not\Leftrightarrow y+z \in A$ (why?). Let
$$
A = \bigcup_{i=0}^\infty \{4^i+1,\ldots,2\cdot 4^i\}.
$$
Given $x &lt; y$, let $z = 4^y + 1 - y$. Then $z + y = 4^y + 1 \in A$, while $x + z &lt; 4^y + 1$ and $x + z \geq z = 4^y + 1 - y &gt; 2 \cdot 4^{y-1}$ (check!), showing that $x + z \notin A$.