Prove that a singleton simply ordered set has a largest element. (James R. Munkres "Topology 2nd Edition") I am reading "Topology 2nd Edition" by James R. Munkres.
There is the following exercise on p.44 in this book:

Let $A$ be a nonempty finite simply ordered set.
(a) Show that $A$ has a largest element. [Hint: Proceed by induction on the cardinality of $A$.]

I proved that a singleton simply ordered set $A$ has a largest element(base case) as follows:

Because $A$ is a singleton set, $A = \{a\}$ for some $a$.
Obviously $a \in \{a\}$.
If $b \in \{a\}$, then $b = a$. So, $b \leq a$. So, $a$ is the largest element of $A$.

Is this proof ok?
Or, I need to prove as follows?

Let $A$ be a singleton simply ordered set.  By definition in this book, this means that there exists a bijection $f$ from $A$ to $S_{1+1} := \{x \in \mathbb{Z}_{+} \mid x < 1 + 1\}$.
I already know that $1 \in S_{1+1}$ and if $n \ne 1$, then $n \notin S_{1+1}$.
Let $a := f^{-1}(1)$. Then $a \in A$.


Assume that $b \neq a$ and $b \in A$.
Then there exists $n\in S_{1+1}$ such that $f^{-1}(n) = b$.
This $n$ must be equal to $1$.
So, $f^{-1}(1) = a \neq b = f^{-1}(1)$.
So $f^{-1}$ is not a mapping.
This is a contradiction.


So $a \in A$ and if $b \neq a$, then $b \notin A$.


So,
$a \in A$.
If $b \in A$, then $b = a$. So, $b \leq a$. So, $a$ is the largest element of $A$.

 A: Let $(A,<)$ be a non-empty finite set with a linear order. This means there is a bijection $f: A \to \{1, 2, \ldots ,n\}$ for some $n \in \Bbb Z^+$. In the text it is shown that $n$ is uniquely determined by $A$ and is called the cardinality of $A$.
Let $$S=\{n \in \Bbb Z^+ \mid \text{ for every linear order of cardinality } n: \max(A) \in A \text{ exists}\}$$ and we want to show it to be an inductive set.
$1\in S$ because if $f: A \to \{1\}$ is a bijection, $A$ indeed has a unique element $a_0$ (which maps to $1$) and this is the maximum (as indeed it's the only element and $a_0 \le a_0$). You're making too much of that part IMO. It's clear that $A= \{f^{-1}(1)\}$ and it's equally clear that any linear order with one element has a maximum.
Suppose $n \in S$ and let $A$ be a linearly ordered set of cardinality $n+1$ so we have a bijection $f: A \to \{1,\ldots,n,n+1\}$. Then the set $A' = f^{-1}[\{1,\ldots,n\}]$ with the restricted order $\le'$ inherited from $A$ has cardinality $n$ ($f\restriction_{A'}$ is a bijection of $A'$ with $\{1,\ldots, n\}$) and as $n \in S$ by assumption, there is $a_0 \in A'$ that is the maximum of $A'$ w.r.t. $\le'$.
Now consider $a_1 = f^{-1}(n+1) (\neq a_0)$: if $a_0 < a_1$ then I claim that $a_1 = \max(A)$: let $a \in A$. If $f(a)=n+1$ then $a = a_1$ and $a \le a_1$ is trivial.
Otherwise, $f(a) < n+1$ so $a \in A'$ and thus $a \le a_0$, and so by transitivity, $a < a_1$ as well. So if $a_0 < a_1$, $\max(A)$ exists. The other case is $a_1 < a_0$, and then I claim $a_0 = \max(A)$: let $a \in A$. If $f(a)=n+1$ then $a=a_1 < a_0$ and we're done. Otherwise $f(a) \le n$ so $a \in A'$ and $a \le a_0$. In either case $a \le a_0$ and $a_0=\max(A)$ exists.
So $S$ is inductive and so $S=\Bbb Z^+$ and it follows that every finite linearly ordered set has a maximum.
