Let $A$ and $B$ be two sets such that $B \subset A$ and there is an injection $f \colon A \to B$. Then how to show that $A$ and $B$ have the same cardinality?
Munkres' Hint:
We define $A_1 \colon= A$, $B_1 \colon= B$, and, for $n > 1$, we define $A_n \colon= f(A_{n-1})$ and $B_n \colon= f(B_{n-1})$.
Thus, we have $$B_1 = B \subset A_1 = A$$ and $$A_2 = f(A_1) = f(A) \subset B = B_1,$$ that is, $$A_2 \subset B_1 \subset A_1.$$ Now if, for any +ive integer $n \geq 1$, we have $$B_{n+1} \subset B_n \subset A_n, $$ then we also have $$B_{n+2} = f(B_{n+1}) \subset f(B_n) = B_{n+1}$$ and $$B_{n+1} = f(B_n) \subset f(A_n) = A_{n+1}, $$ that is, $$B_{n+2} \subset B_{n+1} \subset A_{n+1}.$$ Hence using induction we can conclude that $$A_{n+1} \subset B_n \subset A_n \ \mbox{ for each } \ n \in \mathbb{N}.$$ That is, $$A_1 \supset B_1 \supset A_2 \supset B_2 \supset A_3 \supset B_3 \supset \ldots.$$
Now let the map $h \colon A \to B$ be defined as follows: $$ h(x) \colon= \begin{cases} f(x) \ & \mbox{ if } \ x \in A_n - B_n \ \mbox{ for some } \ n; \\ x \ & \mbox{ otherwise}. \end{cases} $$ Munkres claims that this map $h$ is a bijection. How do we show this?
Injectivity of $h$:
Suppose $a, x \in A$ such that $h(a) = h(x)$. We need to show that $a=x$.
If there exist $m, n \in \mathbb{N}$ such that $a \in A_m - B_m$ and $x \in A_n - B_n$, then we must have $$f(a) = h(a) = h(b) = f(b),$$ which implies $a = b$.
On the other hand, if $a \not\in A_n - B_n$ for any $n \in \mathbb{N}$ and $x \not\in A_n - B_n$ for any $n \in \mathbb{N}$, then we have $$a = h(a) = h(b) = b.$$
How to show that $a = b$ in the following situation?
There exists some $m \in \mathbb{N}$ such that $a \in A_m - B_m$ but $x \not\in A_n - B_n$ for any $n \in \mathbb{N}$.
Surjectivity of $h$:
Let $b \in B$.
Then $b \in A$.
If $b \not\in A_n - B_n$ for any $n \in \mathbb{N}$, then we have $b = h(b)$.
On the other hand, if, for some $k \in \mathbb{N}$, we have $b \in A_k - B_k$, then let $k$ be the smallest such positive integer.
If $k = 1$, then we have $b \in A - B$, which is a contradiction as $b \in B$, by our hypothesis. So we must have $k > 1$; rather, $k \geq 2$.
As $b \in A_k - B_k$ and as $A_k \subset B_{k-1}$, so we must have $b \in B_{k-1}$. What next?
Since you asked me to answer here: http://dbfin.com/topology/munkres/chapter-1/section-7-countable-and-uncountable-sets/problem-6-solution/#comment-2390104583.
Step 1. We show the inclusions $A_1\supset B_1\supset A_2\supset B_2\supset\cdots$, you have already shown it in the question.
Step 2. We show that $h$ is well-defined. Indeed, for all $x\in A$, $h(x)\in B$, as if $x\in A−B=A_1−B_1$ then $h(x)=f(x)\in B$, otherwise $x\in B$, and $h(x)\in \{x,f(x)\}\subset B$, so that $h(x)\in B$.
Step 3. We show that $h$ is bijective. We split all of $A$ into "slices" $C_n=A_n−B_n$ and the rest $X=A−\cup_{n\in\mathbb{Z}_+}C_n$.
(a) All slices and $X$ are pairwise disjoint. $X$ is disjoint from any $C_n$ by definition. Further, for $m>n$, we have $C_m\subset A_m$ (by definition), $A_m\subset B_n$ by Step 1, $B_n\cap C_n=\emptyset$ by definition. Overall, we have $C_m\cap C_n=\emptyset$.
(b) The restriction of $h$ to $C_n$ is a bijection $h_n:C_n\rightarrow C_{n+1}$ because $f|A_n:A_n\rightarrow A_{n+1}$ and $f|B_n:B_n\rightarrow B_{n+1}$ where $B_k\subset A_k$ are bijections.
(c) The restriction of $h$ to X is the identity function.
Since the image sets of the functions $h_n$ and $h|X$ are pairwise disjoint (by (a), (b) and (c)), $A=X\cup_{n\in \mathbb{Z}_+}C_n$ and $h(A)=X\cup_{n\in \mathbb{Z}_+}C_{n+1}=A−C_1=B$, we conclude that $h$ is injective and surjective, i.e. bijective.