I don't understand how "$\mu(X_\infty) \le \int_X f d\mu < \infty$" proves that $f$ is finite a.e. on $X$. I think that this just shows $\mu(X_\infty) < \infty$.
I also don't understand the proof for $\sigma$-finite. By Chebychev's inequality, $\mu(X_n) \le n \cdot \int_X fd\mu < \infty$. But what if $n = \infty$? I think that the inequality does not hold in this case.
I guess $E_n = X_n$ or $X_n \setminus [\cup_{k=1}^{n-1}X_k]$, right?
I appreciate if you elaborate on this.
I agree with you that the proof that $f$ is finite a.e. is incomplete. We really want to show that $\mu(X_\infty) = 0$, not just that $\mu(X_\infty)<\infty$. Notice that $$ X_\infty = \bigcap_{n=1}^\infty \{x\in X : f(x) \ge n\}, $$ and observe that $f(x) \ge n$ for each natural number $n$ and each $x\in X_\infty$. If $\mu(X_\infty) > 0$, then $$ \int_X f\,d\mu \ge \int_{X_\infty}f\,d\mu \ge \int_{X_\infty} n\,d\mu = n\cdot\mu(X_\infty) > 0. $$ Thus $\int_X f\,d\mu = \infty$, a contradiction of what we assumed that $\int_X f\,d\mu < \infty$.
For the second question about the use of Chebychev's inequality, note that Royden says "Let $n$ be a natural number." Thus, $n$ cannot be $\infty$, which is not a natural number.
P.S. Be careful with Royden's book. There are a lot of typos or outright errors in it. I would recommend you read it with a copy of the errata next to you.