You got the definiion of peak slightly wrong. We say that the sequence $(a_n)_{n\in\mathbb N}$ has a peak at index $i\in\mathbb N$, if none of the subsequent members exceeds the value there, i.e. if $a_j\le a_i$ for all $j>i$. We do not exclude, however, that earlier members exceed it.
With this definition it should be clear that infinitely many peaks form a (not necessarily strictly) decreasing subsequence; and if there are only finitely many peaks one can always find (for all positions after the last peak) a next sequence member that exceeds the current one, i.e. one finds a (strictly) increasing subsequence.
An example with infinitely many peaks:
$$ 1, -1, \frac12, -\frac12, \frac13,-\frac13,\frac14,-\frac14,\ldots$$
Here all positive membes are in fact peaks because all later members are smaller.
An example with finitely many peaks:
$$ 42, 17, 33, 0, -\frac12,-\frac13,-\frac14,-\frac15,-\frac15,-\frac16,\ldots$$
Here, the $42$, the $33$ and the $0$ are peaks.
Proving that every sequence has a monotone subsequence isn't a proof that every unbounded sequence has an unbounded monotone subsequence (which it does, but you don't need your subsequence to be monotonic - see below).
Your second approach "for all $k \in \mathbb{N}$ there will exist a $n \in \mathbb{N}$ such that $|x_{n_{k}}| \ge k$"is good, and follows immediately by contradiction as you suggest: if not true for some $k$ then $(x_n)$ is bounded with $|x_n| \le k$.
To generate an unbounded subsequence you pick $n_1$ such that $|x_{n_{k}}| > 1$ then recursively define $k+1$ from $k$ by $|x_{n_{k+1}}| > k+1$ and $n_{k+1} \gt n_k$. The second part of the condition is necessary so that your subsequence has the same order of terms as the original sequence. That such an $x_{n_{k+1}}$ exists follows from the same boundedness argument applied to the tail of the sequence: if not, then the tail is bounded and there are a finite number of (finite) terms up to $n_k$ therefore the sequence $(x_n)$ would be bounded.
Now you have a (not necessarily monotonic) unbounded subsequence with all terms $\ne 0$ so you can form the sequence of inverse terms and clearly for $n_k$ with $k> M$ then all $1/|x_{n_{k}}| \lt 1/M$. So the sequence converges to $0$.
Best Answer
The first statement is false. For $n\in\Bbb N$ let $a_{2n}=0$ and $a_{2n+1}=-\frac1{2n+1}$. Then the sequence
$$\langle a_n:n\in\Bbb N\rangle=\left\langle 0,-1,0,-\frac13,0,-\frac15,\ldots\right\rangle$$
has a peak at each even index, but the subsequence $\langle a_{2n+1}:n\in\Bbb N\rangle$ is strictly increasing.
The second statement is also false: for $n\in\Bbb N$ let $a_n=(-1)^nn$, so that
$$\langle a_n:n\in\Bbb N\rangle=\langle 0,-1,2,-3,4,-5,\ldots\rangle\;;$$
this sequence clearly has a monotonically strictly decreasing subsequence, and it has no peaks at all.