Question about an inductive step Below is a proof of Bolzano-Weierstrass by induction.
Let $A$ be an infinite bounded sequence of reals. Then $A$ has a limit point $l$. Let $a \in A$. Then $\exists a_i \in N_{r}(l)$ for all $r > 0$. In other words, $0< |l - a| < r$ for all $r > 0$.
In particular,  $0< |l - a| < 1$ holds meaning there must be some index $k_1$ for which $0< |l - a_{k_1}| < 1$.
Suppose $0< |l - a_{k_j}| < \frac 1j$ for all $j:= 2, 3, \ldots, n.$
By definition of $l$, we also have $0< |l - a_{k_j}| < \frac {1}{n+1}$ for $j \in \{2,\ldots, n\}$. Because $|N_r(l)| = \infty$ for any $r > 0, \ 0< |l - a_{k_j}| < \frac {1}{n+1}$ holds for infinite number of $k_j$. So far, out hypothesis asserts $0< |l - a_{k_j}| < \frac {1}{n+1}$ is true only for finite number of $k_j$. Thus there's some $k_{n+1}$ s.t. $|l - a_{k_{n+1}}| < \frac {1}{n+1}$. By well-ordering property, we can also assert that $k_{n+1}$ must be the least element of $\{k_m\}_{m=1}^\infty \setminus \{k_1, k_2, \ldots, k_n\}$ that follows after $k_n$.
My question: does the justification above for the existence of $k_{n+1}$ work? In general, does that proof above make sense?
 A: I see multiple problems with your proof. Let me go through it.
The claim you are proving is the following one:

Let $A$ be an infinite bounded sequence of reals. Then $A$ has a limit point $l$. Let $a \in A$. Then $\exists a_i \in N_{r}(l)$ for all $r > 0$. In other words, $0< |l - a| < r$ for all $r > 0$.

There are some problems though. First, there's a typo and $|l-a|$ should be $l-a_i$. Also, you do not seem to be using $a\in A$ at all in that statement. Also saying $0<|l-a|<r$ a for all $r>0$ is a bit problematic as one might confuse it with the impossible statement that the number $|l-a|$ is a non zero number smaller than any positive number. I would therefore rewrite the claim as
Claim: Let $A$ be an infinite bounded sequence of reals, and let $l$ be a limit point for $A$. Namely, for every $r>0$ there exists $a_i\in A$ so that $a_i\in N_r(l)$. In other words, $0<|l-a_i|<r$.

In particular,  $0< |l - a| < 1$ holds meaning there must be some index $k_1$ for which $0< |l - a_{k_1}| < 1$.

So far so good. I would add however that you can write $A$ as a sequence $\{a_1,a_2,...\}$.

Suppose $0< |l - a_{k_j}| < \frac 1j$ for all $j:= 2, 3, \ldots, n.$

I would write, fix $n\geq 1$ and assume inductively that there exists $a_{k_j}\in A$, $j=1,2,...,n$ so that $0<|l-a_{k_j}|<\frac{1}{j}$. Since here you use a claim by induction it may make sense to include another assumption. Namely there exists such $a_{k_j}$'s and $k_1<k_2<...<k_n$.

By definition of $l$, we also have $0< |l - a_{k_j}| < \frac {1}{n+1}$ for $j \in \{2,\ldots, n\}$.

That does not seem to be true, how did you come up with that?

Because $|N_r(l)| = \infty$ for any $r > 0, \ 0< |l - a_{k_j}| < \frac {1}{n+1}$ holds for infinite number of $k_j$.

This sentence is problematic because we have only defined a finite sequence $k_j$. What you really want to write is: Because $|N_r(l)|=\infty$ for any $r>0$, taking $r=\frac{1}{n+1}$ we have that $0<|l-a|<\frac{1}{n+1}$ holds for infinitely many $a\in A$. Note however that the fact that $|N_r(l)|=\infty$ may require a proof.

So far, out hypothesis asserts $0< |l - a_{k_j}| < \frac {1}{n+1}$ is true only for finite number of $k_j$. Thus there's some $n_{k+1}$ s.t. $|l - a_{k_{n+1}}| < \frac {1}{n+1}$.

I would delete the first sentence, and change the second to: Thus there's some $k_{n+1}$ s.t. $0<|l-a_{k_{n+1}}|<\frac{1}{n+1}$.

By well-ordering property, we can also assert that $k_{n+1}$ must be the least element of $\{k_m\}_{m=1}^\infty \setminus \{k_1, k_2, \ldots, k_n\}$ that follows after $k_n$.

You can use it, but since I have assumed that $k_1<k_2<...<k_n$ you no longer have to (note that you need that anyway). Instead you can say that since $N_r(l)$ is infinite for $r=\frac{1}{n+1}$ and $\{a_1,a_2,...,a_{k_n}\}$ is finite, there must exists $k_{n+1}>k_n$ so that $a_{k_{n+1}}\in N_{\frac{1}{n+1}}(l)$.
