Show that $z$ is a cluster point of a sequence if and only if there is a subsequence $\{x_{n_k}\}$ converging to z Note: a cluster point (in a sequence) is defined as a number z with a property that
$\forall \epsilon > 0,$ there are infinitely many terms of the sequence in the interval
$(z-\epsilon,z+\epsilon)$

How do I go about starting this one? This just sounds like the definition of limit to me. Right? Because with a given limit $L$ there are also infinitely many terms in $(L-\epsilon,L)$

Obviously, the fact that we'll be showing that there is a subsequence that converges to this cluster point shows the relationship between limits and this cluster point.

($\Rightarrow$): Suppose $\exists$ subsequence $\{x_{n_k}\}$ with $\{x_{n_k}\}$ converging to z. $\Rightarrow N \in \mathbb{N} :|x_{n_k}-z|<\epsilon \Rightarrow |x_{n_k}-z|<\epsilon \Rightarrow$ $$\lim_{k\to\infty}  x_{n_k}=z$$ $\Rightarrow z$ is a cluster point

($\Leftarrow$): The same as the first direction, but in reverse
 A: 
Suppose $\exists$ subsequence ${x_{n_k}}$ with ${x_{n_k}}$ converging to $z$.

Why mix in symbols from first-order logic into an English sentence? Just use English to write statements and then use symbols as and when it makes sense to.

$N \in \mathbb{N}: |x_{n_k} -z| < \epsilon \implies |x_{n_k}-z| < \epsilon$

I mean, sure? Don't know what the purpose of writing that down is but yea. Also, what role does $N$ play? And besides, just saying that:
$$\lim_{k \to \infty} x_{n_k} = z$$
doesn't prove that $z$ is a cluster point. You haven't actually provided any argument for that claim.

We start by proving the forward direction first. Suppose that $z$ is a cluster point of the sequence $\{x_n\}$. Then, for every $\epsilon > 0$, there are infinitely many terms of the sequence in the interval $(z-\epsilon,z+\epsilon)$.
Define a subsequence $\{x_{n_k}\}$ inductively. If we have the following set:
$$A = \{n \in \mathbb{N}:|x_n-z| < \frac{1}{n} \}$$
Then, define $x_{n_1} := x_{\min(A)}$. We know that $\min(A)$ exists by the well-ordering property of $\mathbb{N}$. Once $x_{n_k}$ has been defined, we consider the following set:
$$B = \{n \in \mathbb{N}: |x_n-z| < \frac{1}{n} \land n > n_k\}$$
Then, define $x_{n_{k+1}} := x_{\min(B)}$. We can then use the squeeze theorem to conclude that this subsequence converges to $z$.
For the backwards direction, we suppose that there is a subsequence of $\{x_n\}$, given by $\{x_{n_k}\}$ such that:
$$\lim_{k \to \infty} x_{n_k} = z$$
Then, this is equivalent to saying that:
$$\forall \epsilon > 0: \exists K \in \mathbb{N}: \forall k \geq K: |x_{n_k}-z| < \epsilon$$
Let $\epsilon > 0$ be arbitrary but fixed. Then, that fixes $K \in \mathbb{N}$ such that the above holds. Since the following set is finite:
$$\{n \in \mathbb{N}: n < K\}$$
it follows that the set $\{n \in \mathbb{N}: n \geq K \}$ is infinite. That means that there are infinitely many terms of the subsequence, and therefore of the sequence, in the interval $(x-\epsilon,x+\epsilon)$. That proves the desired result.
