A form of the Peter–Weyl theorem says that
irreducible representations of compact groups are finite-dimensional.
I know that the same statement is not true in case of compact semigroups. But I don’t know why it is not true. Any insight and/or an example of a compact semigroup with infinite-dimensional irreducible representation would be great.
There is of course continuity to assume, otherwise compactness has no sense; let us suppose that the representations should be weakly continuous.
If you allow the semigroup to be just semitopological, that is, its multiplication to be separately continuous, then one can take the unit ball in $B(H)$ – the space of bounded linear operators on a Hilbert space – in the weak operator topology. It is compact, and its natural representation on the same space $H$ is of course irreducible; $H$ can be taken infinite-dimensional.
If you want the multiplication to be jointly continuous, that is to have a topological semigroup, then the space matters. On a Banach space one can find even a commutative example. Take $V = l_1$, and let $T$ be a bounded operator on $V$ without nontrivial invariant subspaces (examples of Enflo and then of Read). The map $\pi(x) = \exp(Tx)$ is an irreducible representation of the real line on $V$, so restricted on $[0,1]$ it remains irreducible and your semigroup is compact.
For a Hilbert space I cannot think of an example at the moment.
But the absence of Peter–Weyl is not only in the case there are infinite-dimensional irreducibles, it is also in the case when finite-dimensional representations do not separate points. And such an example is easy to find: take $[0,1]$ with the usual topology and the operation $s \wedge t= \min(s,t)$, then its only nonzero finite-dimensional representation is the constant one, that is, every $\pi(s)$ is the same projection (since every $\pi(s)$ is a projection, and moreover the image of every $\pi(s)$ must be invariant under any other $\pi(t)$). Note also that every nonconstant representation of this semigroup is reducible.