Confused regarding Terence Tao's proof of Tonelli's Theorem I am struggling with Tao's proof of Tonelli's Theorem perhaps, admittedly, due to not having properly studied all the previous material regarding product measures in his book on the topic. I'd appreciate any help understanding any of the steps in the proof I am confused by.

Theorem 1.7.15 (Tonelli’s theorem, incomplete version). Let $(X, B_X, \mu_X)$ and $(Y, B_Y , \mu_Y )$ be $\sigma$-finite measure spaces, and let $f : X \times Y \to [0, +\infty]$ be measurable with respect to $B_X \times B_Y$. Then:
(i) The functions $x\mapsto \int_Y f(x,y)d\mu_Y$ and $y\mapsto \int _X f(x,y) d\mu_X$ are measurable with respect to $B_X$ and $B_Y$ respectively.
(ii) We have
$$\int_{X\times Y} f(x,y) \ d(\mu_X\times\mu_Y) = \int_Y\left(\int_X f(x,y) \ d\mu_X\right)d\mu_Y.$$
Proof: By writing the $\sigma$-finite space $X$ as an increasing union $X = \cup_{n=1}^{\infty} X_n$ of finite measure sets, we see from several applications of the monotone convergence theorem (Theorem 1.4.44) that it suffices to prove the claims with $X$ replaced by $X_n$. Thus we may assume without loss of generality that $X$ has finite measure. Similarly we may assume $Y$ has finite measure. Note from (1.36) that this implies that $X × Y$ has finite measure also.

Q1: why does it suffice to show the result for $X$ replaced by $X_n$? I do not even understand how replacing $X$ with $X_n$ allows us to use the monotone convergence theorem. However, letting $X'_n=\cup_{i=1}^nX_i$ and, $Y'_n=\cup_{i=1}^nY_i$ we get
$$f = \lim_{n\to \infty}f\ 1_{X'_n\times Y'n},$$
and the monotone convergence theorem can be applied, yet -if this is what Tao means- how may we use the monotone convergence theorem to conclude the result holds for $f$ given that it holds for any $f\ 1_{X'_n\times Y'n}$?

Every unsigned measurable function is the increasing limit of unsigned simple functions. By several applications of the monotone convergence theorem (Theorem 1.4.44), we thus see that it suffices to verify the claim when $f$ is a simple function. By linearity, it then suffices to verify the claim when $f$ is an indicator function, thus $f = 1_S$ for some $S \in B_X × B_Y$.

Q2: why is every unsigned measurable function the increasing limit of simple unsigned measurable functions?

Let $C$ be the set of all $S \in B_X × B_Y$ for which the claims hold. From the repeated applications of the monotone convergence theorem (Theorem 1.4.44) and the downward monotone convergence theorem (which is available in this finite measure setting) we see that C is a monotone class.

Q3: Let $A_1\subseteq A_2 \subseteq \ldots $ all be members of $C$, let $A'_n = \cup_{i=1}^nA_n$ and $A = \cup_{i=1}^{\infty}A_n$. Since
$$1_A = \lim_{n\to \infty}1_{A'_n}$$
we may apply the monotone convergence theorem to conclude that $C$ is closed under countable monotone unions. However, I do know how to show that $C$ is closed under countable monotone intersections.

By direct computation (using (1.36)), we see that $C$ contains as
an element any product $S = E × F$ with $E \in B_X$ and $F \in B_Y$. By finite additivity, we conclude that $C$ also contains as an element any a disjoint finite union $S = E_1×F_1∪\ldots ∪E_k×F_k$ of such products. This implies that $C$ also contains the Boolean algebra $B_0$ in the proof of Proposition 1.7.11, as such sets can always be expressed as the disjoint finite union of Cartesian products of measurable sets. Applying the monotone class lemma, we conclude that $C$ contains $\langle B_0\rangle = B_X × B_Y$, and the claim follows.

Q4: The algebra $B_0$ is the collection of all finite unions
$$E_1\times F_1 \cup \ldots \cup E_n\times F_n$$ of $B_X$-measurable sets $E_k$ and $B_Y$-measurable sets $F_k$. How does one show $B_0$ is closed under complements and finite intersections?

 A: I must say that you probably could have answered all your questions by yourself if you read the chapters before the theorem as it is a really good book on measure theory :)
Nevertheless, let me try to answer your questions.

Q1: why does it suffice to show the result for $X$ replaced by $X_n$? I do not even understand how replacing $X$ with $X_n$ allows us to use the monotone convergence theorem. However, letting $X'_n=\cup_{i=1}^nX_i$ and, $Y'_n=\cup_{i=1}^nY_i$ we get
$$f = \lim_{n\to \infty}f\ 1_{X'_n\times Y'n},$$
and the monotone convergence theorem can be applied, yet -if this is what Tao means- how may we use the monotone convergence theorem to conclude the result holds for $f$ given that it holds for any $f\ 1_{X'_n\times Y'n}$?

Suppose the theorem holds for finite measure spaces.
By the definition of $\sigma$-finiteness there exist measurable $X_n$, $Y_n$ such that
$$
X = \bigcup_{n \in \mathbb{N}} X_n, \\
Y = \bigcup_{n \in \mathbb{N}} Y_n,
$$ with $\mu_X(X_n), \mu_Y(Y_N) <\infty$ for all $n \in \mathbb{N}$.
Of course we can assume them to be increasing, i.e. $X_n \subseteq X_{n+1}, Y_n \subseteq Y_{n+1}$ for all $n \in \mathbb{N}$.
It follows that
$$
f = \lim_{n \to \infty} f\cdot\mathbb{1}_{X_n \times Y_n}
$$ pointwise (and in an increasing manner).
Therefore, by the monotone convergence theorem,
$$
\int_{X \times Y} \, f(x, y) \, (\mu_X \otimes \mu_Y)(\mathrm{d}x, \mathrm{d}y)
= \int_{X \times Y} \, \lim_{n \to \infty} f(x, y) \cdot \mathbb{1}_{X_n \times Y_n}(x, y) \, (\mu_X \otimes \mu_Y)(\mathrm{d}x, \mathrm{d}y)
\\
= \lim_{n \to \infty} \int_{X \times Y} \,f(x, y) \cdot \mathbb{1}_{X_n \times Y_n}(x, y) \, (\mu_X \otimes \mu_Y)(\mathrm{d}x, \mathrm{d}y)
= \lim_{n \to \infty} \int_{X_n \times Y_n} \, f_n(x, y) \, (\mu_{X_n} \otimes \mu_{Y_n})(\mathrm{d}x, \mathrm{d}y) 
\\
= \lim_{n \to \infty} \int_{X_n}  \, \left(\int_{Y_n} \, f_n(x, y) \, \mu_{Y_n}(\mathrm{d}y) \right) \, \mu_{X_n}(\mathrm{d}x)
= \lim_{n \to \infty} \int_{X} \, \mathbb{1}_{X_n}  \, \left(\int_{Y} \, f(x, y) \, \mathbb{1}_{Y_n} \, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x)
\\
= \int_{X} \, \lim_{n \to \infty} \mathbb{1}_{X_n}  \, \left(\int_{Y} \, f(x, y) \, \mathbb{1}_{Y_n} \, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x)
= \int_{X} \, (\lim_{n \to \infty} \mathbb{1}_{X_n})  \, \left(\int_{Y} \, \lim_{n \to \infty}  f(x, y) \, \mathbb{1}_{Y_n} \, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x)
\\ 
=  \int_{X} \, \left(\int_{Y} \, f(x, y) \, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x)
$$ where $f_n$ is just the restriction of $f$ to $X_n \times Y_n$ and we used that the theorem is supposed to hold for finite measure spaces.
Notice that the restriction of $\mu_{X} \otimes \mu_Y$ to $X_n \times Y_n$ is just the product measure $\mu_{X_n} \otimes \mu_{Y_n}$ by uniqueness of the product measure(see Proposition 1.7.11).

Q2: why is every unsigned measurable function the increasing limit of simple unsigned measurable functions?

What is your definition of (unsigned) measurable function and how do you define integrals w.r.t. a measure of them?
This is a really basic fact of measure theory and some people even define measurable functions as such.
Nevertheless, a standard construction is something along the lines of
$$
f = \lim_{n \to \infty} \left( \sum_{k = 0}^{n\cdot2^n - 1} k\cdot2^{-n}\cdot \mathbb{1}_{k\cdot2^{-n} \leq f < (k+1)\cdot2^{-n}} + n\cdot\mathbb{1}_{f \geq n} \right).
$$

Q3: Let $A_1\subseteq A_2 \subseteq \ldots $ all be members of $C$, let $A'_n = \cup_{i=1}^nA_n$ and $A = \cup_{i=1}^{\infty}A_n$. Since
$$1_A = \lim_{n\to \infty}1_{A'_n}$$
we may apply the monotone convergence theorem to conclude that $C$ is closed under countable monotone unions. However, I do know how to show that $C$ is closed under countable monotone intersections.

The same idea works for a decreasing sequence of sets.
Let $A_1 \supseteq A_2 \supseteq \dots $ all be elements of $\mathcal{C}$ and define $A := \bigcap_{n \in \mathbb{N}} A_n$.
Then $\mathbb{1}_A$ is the pointwise limit of the decreasing sequence of functions $\mathbb{1}_{A_n}$ and we can deduce
$$
\int_{X \times Y} \, \mathbb{1}_{A}(x, y) \, (\mu_X \otimes \mu_Y)(\mathrm{d}x, \mathrm{d}y)
= \int_{X \times Y} \, \lim_{n \to \infty} \mathbb{1}_{A_n}(x, y) \, (\mu_X \otimes \mu_Y)(\mathrm{d}x, \mathrm{d}y)
\\
=  \lim_{n \to \infty} \int_{X \times Y} \, \mathbb{1}_{A_n}(x, y) \, (\mu_X \otimes \mu_Y)(\mathrm{d}x, \mathrm{d}y)
= \lim_{n \to \infty} \int_{X} \, \left(\int_{Y} \, \mathbb{1}_{A_n}(x, y)\, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x)
\\
= \int_{X} \, \left(\int_{Y} \,  \lim_{n \to \infty} \mathbb{1}_{A_n}(x, y)\, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x)
= \int_{X} \, \left(\int_{Y} \, \mathbb{1}_{A}(x, y)\, \mu_{Y}(\mathrm{d}y) \right) \, \mu_{X}(\mathrm{d}x).
$$
The interchange of limits and integrals is legal because $X, Y$ have finite measure; we could just apply the upwards monotone convergence theorem to $1 - \mathbb{1}_{A_n}$ which converges to $1-\mathbb{1}_A$ pointwise and increasing.
Therefore, $A \in \mathcal{C}$ too.

Q4: The algebra $B_0$ is the collection of all finite unions
$$E_1\times F_1 \cup \ldots \cup E_n\times F_n$$ of $B_X$-measurable sets $E_k$ and $B_Y$-measurable sets $F_k$. How does one show $B_0$ is closed under complements and finite intersections?

This is also a pretty standard exercise in introductions to measure theory.
Nevertheless, let $E_1, \dots, E_n, \tilde{E}_1, \dots, \tilde{E}_m \in \mathcal{B}_X$ and $F_1, \dots, F_n, \tilde{F}_1, \dots, \tilde{F}_m \in \mathcal{B}_Y$.
Then we have
$$
\left(\bigcup_{j = 1}^{m} E_j \times F_j \right) \cap \left(\bigcup_{j = 1}^{m} \tilde{E}_j \times \tilde{F}_j \right)
= \bigcup_{j = 1}^{m} \left( (E_j \times F_j) \cap (\tilde{E}_j \times \tilde{F}_j)  \right)
= \bigcup_{j = 1}^{m} (E_j \cap\tilde{E}_j) \times (F_j \cap\tilde{F}_j)
$$
and
$$
\left(\bigcup_{j = 1}^{m} E_j \times F_j \right)^C
= \bigcap_{j =1}^{m} (E_j \times F_j)^C
= \bigcap_{j =1}^{m} \left((E^C \times F) \cup (E \times F^C) \cup (E^C \times F^C) \right)
$$
Why does this suffice?
