Converse of Bolzano-Weierstrass theorem

Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true.

That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.

Here's the example give in the book: The set: {1, 2, 1, 4, 1, 6, ...} is unbounded, but has a limit point of 1. I can't understand how this set has a limit point as 1. According to the book definition of limit point, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. If I apply it here, then I get only infinity as the limit point. Am I missing something?

It seems there is some confusion between sets and sequences. In this example, what you have written as "{1, 2, 1, 4, 1, 6, ...}" is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,\dots$.

In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $n\in\mathbb{N}$ such that $a_n\in U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,\dots$, and so $1$ is a limit point of the sequence.

(In contrast, $1$ is not a limit point of the set $\{1, 2, 1, 4, 1, 6, \dots\}=\{1, 2, 4, 6, \dots\}$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)

We have

$$a_n = \begin{cases} 1, & n \text{ is odd} \\ n, & n \text{ is even}\end{cases}$$

$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.