Here is one proof that I found in Section 1.8 of James Norris' Markov Chains. It's far more sophisticated than I thought would be necessary; if anybody has a more straightforward proof to offer it would be much appreciated.
Suppose $X_n$ is null recurrent. If it is periodic with period $d$, we can consider the aperiodic subchain $X_{dn}$ and prove that return probabilities for this chain tend to zero, thereby proving that all return probabilities tend to zero. Hence assume the chain is aperiodic.
Let $T_j$ be the first return time to state $j$. We know that
$$ \infty = \sum_{k = 0}^\infty k \,\mathbb{P}_j(T_j = k) = \sum_{k = 0}^\infty \mathbb{P}_j(T_j > k) \,.$$
Given $\varepsilon > 0$ choose $K$ such that
$$ \sum_{k=0}^K \mathbb{P}_j(T_j > k) \geq \frac{1}{\varepsilon} \,.$$
Then, for $n \geq K$,
$$
\begin{align}1 &\geq \sum_{k =n-K}^n\mathbb{P}(X_k = j, X_m \neq j \;\text{for}\; m=k+1,\ldots,n) \\
&= \sum_{k =n-K}^n \mathbb{P}(X_k = j)\mathbb{P}_j(T_j > n - k) \\
&=\sum_{k =0}^K \mathbb{P}(X_{n-k} = j)\mathbb{P}_j(T_j > k)\,.\end{align}$$
Hence we must have $\mathbb{P}_j(X_{n-k} = j)\leq \varepsilon$ for some $k = 0,1,\ldots,K$.
Now we make a coupling argument. Let $Y_n$ be a chain with the same transition probabilities as $X_n$ and an initial distribution to be defined later. The chain $W_n = (X_n,Y_n)$ is irreducible by virtue of the aperiodicity of $X_n$. If $W_n$ is transient then by setting $X_n$ and $Y_n$ to have the same initial distribution we can easily show that return probabilities tend to zero.
So assume that $W_n$ is recurrent. But an irreducible recurrent chain explores all states with unit probability, so the two chains $X_n$ and $Y_n$ meet almost surely. Set $Z_n$ to equal $X_n$ before the two chains meet (at $n = T$, say) and $Y_n$ thereafter. Then $X_n$ and $Z_n$ have the same distribution so
$$\begin{align}|\mathbb{P}(X_n = j) - \mathbb{P}(Y_n = j)| &= |\mathbb{P}(Z_n = j) - \mathbb{P}(Y_n = j)| \\
&=|\mathbb{P}(X_n = j, n < T) - \mathbb{P}(Y_n = j, n < T)| \\
& \leq \mathbb{P}(n < T) \to 0 \,.
\end{align} $$
Hence the two chains $X_n$ and $Y_n$ have converging return probabilities. Now we just need to exploit the fact that we are free to choose the initial distribution for $Y_n$. If $X_n$ has initial distribution $\lambda$ let $Y_n$ have initial distribution $\lambda P^k$ for $k = 1,\ldots,K$, so that $\mathbb{P}(Y_n = j) = \mathbb{P}(X_{n+k} = j)$. Since the two chains have convergent probabilities, we can find $N$ such that for $n \geq N$ and $k=1,\ldots,K$,
$$ | \mathbb{P}(X_n = j) - \mathbb{P}(X_{n+k} = j)| \leq \varepsilon \,.$$
Finally, we know from before that in any interval of length $K$ there is some $k$ such that $\mathbb{P}(X_k = j) \leq \varepsilon$, and hence
$$ \mathbb{P}(X_n = j) \leq 2 \varepsilon \,.$$
Since $\varepsilon > 0$ was arbitrary we have proved the result.
I think that claim is not trivial. In the book referenced in your link, that statement is observed as a corollary to the ergodic theorem stating that $p_{ij}(n) \to \frac{1}{\mathbb E_x T_x}$ for any irreducible aperiodic Markov chain. You can find a proof of that result here, for instance.
If all you want to prove is your original claim (that all irreducible finite Markov chains are positive recurrent), I think there's an easier way to do it than by that lemma. Assume aperiodicity for simplicity, but periodic chains just make the proof more annoying (rather than prevent the result from being true). The sketch of the proof is:
For an irreducible, aperiodic finite Markov chain, there exists some $m$ such that $p_{ij}(m) > 0$ for all state pairs $\{i, j\}$.
Moreover, because the chain is finite you can strengthen the above to say that $p_{ij}(m) > \epsilon$ for some universal $\epsilon > 0$.
Hence, you can dominate the excursions of $X_n$ from $x$ to $x$ with a geometric random variable by conditioning on the behavior on every $m^{\text{th}}$ step.
Since the geometric random variable has a finite expectation, so do the excursion lengths.
EDIT: Expanding the answer to address the steps to show step 1 above. I'll work with the following definition of aperiodicity: a state $x$ is aperiodic if $\gcd(k:p_{xx}(k) > 0) = 1$. In all the following, I'll also assume irreducibility of the Markov chain. Without irreducibility, some version of these basic ideas all still hold, but they're just more obnoxious to write down; you'd just have to address an irreducible subclass of states.
Lemma: If state $x$ is aperiodic, then there exists $N$ such that $p_{xx}(n) > 0$ for all $n \geq N$.
Corollary: For the aperiodic state $x$ above, there exists $N'$ such that for all other states $y$ and all $n \geq N'$, $p_{xy}(n) > 0$.
Corollary to the corollary: If a chain has one aperiodic state, all states are aperiodic.
Best Answer
Probably a bit late but yes, this can be shown quiet nicely.
For more details just comment below.