NOTE: I know that there are similar questions (1, 2, 3), but none of those are asking for the proof-verification (and this is exactly the point of my question). Furthermore, after sifting through the answers to those questions, I haven't found any answer that provided a proof resembling mine (hence no possibility of self-verification). Therefore, I would like someone to verify my approach.
Couple more notes:
I will denote cardinality of arbitrary set A as $|A|$.
I will use the folowing definition of finiteness: "Set $A$ is finite iff there exists bijection between $A$ and $\{i\}_{i=1}^{n}$ for some natural number $n$."
We will assume that following holds: "If $A$ is finite set, $|A| = n$ ($n \in \mathbb N$ and $n ≥ 1$) and $a \in A$, then $A \setminus \{a\}$ is finite and $|A \setminus \{a\}| = n -1$
Proposition. Let $A$ be arbitrary finite set with $|A| = n$. Show that for arbitrary $B \subseteq A$, $B$ is finite B is finite with $|B| ≤ |A|$.
My attempt:
We will induct on cardinality of $A$.
For the base case, we will consider $n = 0$ and $n=1$.
Take arbitrary set $A$ and suppose $|A| = 0$. Then the only subset of $A$ is $\emptyset$. Clearly, $\emptyset$ is finite with zero cardinality, implying that $|\emptyset| ≤ |A|.$
Now, take arbitrary set $A$ and suppose that $|A| = 1$. $A$ has two subsets: $|\emptyset|$ and $A$ itself. Clearly, $|\emptyset| ≤ |A|$ and $|A| ≤ |A|$.
Now inductively suppose that for arbitrary set with cardinality $n$ (note that $n >1$), its arbitrary subset is finite with cardinality smaller or equal to $n$.
Consider arbitrary set with cardinality $n+1$, say $B$. Take arbitrary $b \in B$ and define $A := B \setminus \{b\}$. We know that $|A| = n$, implying that arbitrary $Z \subseteq A$ is finite with $|Z| ≤ n$.
Now take arbitrary subset of $B$, say $K$. We shall consider three cases:
$b \notin K$
$K = \{b\}$
$b \in K$ and $K ≠ \{b\}$
First case. Suppose $b \notin K$. Then $K \subseteq A$, and by inductive hypothesis we know that $|K| ≤ n < |B|$.
Second case. Suppose $K = \{b\}$. Define $g: \{1\} \rightarrow \{b\}$ with $g(1) = b$. Clearly, $g$ is a bijection, and so $K$ is finite with $|K| = 1$. Hence $|K| ≤ |B|$.
Third case. Suppose $b \in K$. Define $P := K \setminus \{b\}$. It can be verified that $P \subseteq A$, and so $P$ is finite with $|P| ≤ n$. Let $|P| = p$. Since $P$ has cardinality $p$, we can define some bijection $f: \{i\}_{i=1}^{p} \rightarrow P$. Now define function
$$g: \{i\}_{i=1}^{p+1} \rightarrow K$$
such that $g(i) = f(i)$ for $1 ≤ i ≤ p$ and $g(p+1) = b$. It is not hard to check that $g$ is indeed a bijection. Furthermore, note that $p ≤ n < n+1$, hence clearly $p < n+1$ and thus $p + 1 ≤ n + 1$.
Therefore, for arbitrary $K \subseteq B$, regardless of whether $b$ is in $K$ or not, subset $K$ will be finite with a cardinality smaller or equal to the cardinality of $B$. $\Box$
Remark 1. I considered the case $K = \{b\}$ separately (instead of moving it into the third case), because I would run into a problem otherwise: in the third case, I asserted that if we define $P := K \setminus \{b\}$, then it would be possible to define some bijection between $\{1,\cdots,p\}$ and $P$. Clearly, the assertion would fail in case of $K = \{b\}$, because $P = K \setminus \{b\}$ would be an emptyset, and no bijection could be formed.
Q. Is the proof correct?
