I have an infinite set $A$ of real numbers and I wish to construct a strictly increasing sequence using elements of $A$. This is what came to my mind -
Pick any element from $A$ and name it $a_1$. Pick a different element from $A$. If this is greater than $a_1$, call it $a_2$. If not, rename the first element $a_2$ and the new one $a_1$. Suppose I have chosen $a_1<a_2<…<a_n$. Pick an element $a$ different from all these. If $a>a_n$, name it $a_{n+1}$. If $a<a_1$, rename $a_i$ as $a_{i+1}$ for all $i$, and $a$ as $a_1$. If $a_k<a<a_{k+1}$, rename $a$ to $a_{k+1}$ and $a_i$ to $a_{i+1}$ for $i\geq k+1$. This way I have $a_1<a_2<…<a_n<a_{n+1}$. Then by induction I have my increasing sequence $(a_n)$.
Am I allowed to do this? Did I invoke any axiom while doing this?
This proves you can create an increasing sequence of any finite length (because if a sequence of length $n$ can be made, then a sequence of length $n+1$ can be made) but not an increasing sequence of infinite length (because for any $n$ you can "go one more" to length $n+1$ but adding one will never get you to length $\infty$) .
It's a subtle but important point that induction can give you a conclusion for all natural values but not for a countable infinite value.[1]
In this case it will be possible that for an infinite number of attempts we will find a number $<$ than $a_1$ and thus a new $a_1$ will be assigned an infinite number of times and there will be no "final" $a_1$ for the infinite sequence.
As Tony Ks answer shows the set is $\{-1,-2, -3, ....\}$ then the first sequence will be $\{a_1\} = \{-1\}$ and the second sequence will be $\{a_1, a_2\} = -2, -1$, and so on. Then $n$ sequence will be $\{a_1, .... a_n\} = \{-n, -n+1, ...., -1\}$. But each sequence will always have a different $a_1$ and there will be no "agreed upon" $a_1$ for an infinite sequence.
[1] A false proof that all real numbers between $0$ and $1$ is rational: If every decimal with $n$ places is rational then a decimal with $n+1$ spaces $0.a_1a_2....a_na_{n+1} = 0.a_1a_2.....a_n + \frac {a_{n+1}}{10^{n+1}}$ is the sum of two rational numbers and hence rational. So by induction every decimal with any number of places is rational. As every real number between $0$ and $1$ can be written a decimal with some number (possibly infinite) places, every such real number is rational.
The catch is that induction only proves something about decimals with finite number of places. Not infinite number of places. Although we can "add $1$" an infinite number of times to get an infinite number of numbers, those numbers are all finite. We can never "add $1$" and ever get $\infty$.