Is the propositional set countably infinite?

Yes, PROP is countably infinite. It’s clearly infinite, because each of the strings $P_i$ for $i\in\Bbb N$ belongs to PROP. To show that it’s only countably infinite, it suffices to find an injection from PROP to a set that is already known to be countably infinite. One way to do this is to find a countably infinite set that contains PROP as a subset.

Let $S$ be the set whose elements are the proposition symbols $P_i$ for $i\in\Bbb N$ and the symbols $\lnot$ (or ~), $\lor$, $\land$, and the left and right parenthesis symbols. $S$ is a countably infinite set.

This is pretty obvious, but in any case we can easily write down an explicit bijection $\varphi$ between $S$ and $\Bbb N$; one is given by $\varphi(0)=\lnot$, $\varphi(1)=\lor$, $\varphi(2)=\land$, $\varphi(3)=($, $\varphi(4)=)$, and $\varphi(n)=P_{n-5}$ for $n\ge 5$.

Every string in PROP has a length, and that length is a positive integer. If $\sigma\in\text{PROP}$ has length $n$, then $\sigma\in S^n=\underbrace{S\times S\times\ldots\times S}_{n\text{ times}}$. Thus, every string in PROP belongs to the set

$$T=\bigcup_{n\in\Bbb Z^+}S^n=S\cup S^2\cup S^3\cup S^4\cup\ldots\;\;.$$

Since $\text{PROP}\subseteq T$, all that’s necessary in order to show that PROP is countably infinite is to show that $T$ is countably infinite. To do this in detail requires several steps.

1. Prove that $S^2$ is countably infinite; this can be done with the Cantor pairing function.

2. Prove by induction on $n$ that $S^n$ is countably infinite for each $n\in\Bbb Z^+$.

3. Prove that the union of countably many countably infinite sets is countably infinite. This can be done by a slightly more sophisticated use of the pairing function. Once you know that each $S^n$ is countably infinite, you know that each can be listed as $S^n=\{\sigma_n(k):k\in\Bbb N\}$. You then use the pairing function $\pi$ to assign each $\sigma_n(k)\in T$ the natural number $\pi(n,k)$; this gives a bijection from $T$ to $\Bbb N$.

Each strng in PROP is a finite sequence of symbols from a countable alphabet (consisting of $P_0, P_1, \ldots$ and the logical symbols). For each $n$, the set of such sequences of length $n$ can be viewd as subset of $\mathbb N^n$. Since the diagonal argument shows that $|\mathbb N\times \mathbb N|=|\mathbb N|$, ist follows by induction that $|\mathbb N^n|=|\mathbb N|$. By the same diagonal argument we see that PROP as the (here: disjoint) union of countably many countable sets is countable.

Just how quick and dirty a proof is acceptable? Take your favourite Gödel-numbering style coding system for strings of symbols, and take the function that takes $n$ to the wff with code $n$ (if there is one) and to $P_0$ otherwise. And voilà, a surjective function from $\mathbb{N}$ to the set of wffs, which makes the latter countable by definition.