The conditional statement is P(n), given by,
$$P(n):\forall \ n \in \mathbb{N}, \text{if} \ |A|=n \ \text{then} \ A \ \text{is bounded}$$ The base case is when $n=1$. Thus $A=\{a_0\}$ where $a_0 \in \mathbb{R}$. $|\{a_0\}|=1=n$. Therefore, our base case is bounded by $M>1$.\ The inductive assumption. I will assume that $k$ is fixed and that $P(k)$ is true. That is, assume $|A|=k$ and $A$ is bounded. $$$$ The inductive step. Now I will demonstrate that also $P(k+1)$ is true. I must establish that $|A|$ is bounded.There are two cases. Suppose that all sets of cardinality $|k-1|$ are bounded. Let $A$ be any set of size $k$. Take a subcollection of $|k-1|$ elements. We are left with one element and our two cases. The element is k and the cases are when, cardinality of $A$ is equal to k and when the cardinality of $A$ is not equal to k. Below we will examine the two cases, \begin{align*} \text{ Case one:} |A|=k\\ \text{Let} \ B \end{align*} I'm still struggling to figure out how to use cardinality with induction and to determine the two cases Andres mentioned.
Base: $\{a_0\}$ is clearly bounded.
Induction hint:: There are two cases. Suppose that all sets of cardinality $|n-1|$ are bounded. Let $A$ be any set of size $n$. Take a subcollection of $n-1$ elements. There is one element left over and two cases, what are they?