Why do I need to show uniqueness? Question
Let $c$ be a cluster point of $A  \subset \mathbb{R}$, and $f:A \to \mathbb{R}$ be a function. Suppose for every sequence $\{ x_n \}$ in $A$, such that $\lim_{n \to \infty} x_n = c$, the sequence  $\{ f(x_n)\}_{n=1}^{\infty}$ is Cauchy. Prove that $\lim_{x \to c} f(x)$ exists.
What I came up with for Proof:
Note that  $\{ f(x_n)\}$ being Cauchy implies that it is a convergent sequence. Let $\lim_{n \to \infty} f(x_n)=L$
Then by Lemma 3.1.7: $f(x) \to L$ as $x \to c$
Lemma 3.1.7 is in our text book and states:
Let $S  \subset \mathbb{R}$ and $c$ be a cluster point of $S$. Let $f: S \to \mathbb{R}$ be a function.
Then $f(x) \to L$ as $x \to c$ if and only if for every sequence $\{x_n\}$ of number such that $x_n \in S  \setminus \{c\}$ for all $n$ and such that lim $x_n=c$, we have that the sequence $\{f(x_n)\}$ converges to $L$.
My professor mentioned that the first step should be to show the lim f($x_n$) exists. 
Then the second step should be: If ($x_n$) is convergent to c then lim f($y_n$) exists.
Then he said that I need to show that $\lim_{n \to \infty} f(x_n)=\lim_{n \to \infty} f(y_n)$
I don't understand why I would have to do the second two steps
 A: The reason why you need "uniqueness" is stated in your Lemma 3.1.7. (emphasis mine)

Let $S  \subset \mathbb{R}$ and $c$ be a cluster point of $S$. Let $f: S \to \mathbb{R}$ be a function.
Then $f(x) \to L$ as $x \to c$ if and only if for every sequence $\{x_n\}$ of number such that $x_n \in S  \setminus \{c\}$ for all $n$ and such that $\lim x_n=c$, we have that the sequence $\{f(x_n)\}$ converges to $L$.

It could happen that we have two different sequences $\{x_n\}$ and $\{y_n\}$ that both converge to $c$ such that $\lim_{n \rightarrow \infty} f(x_n) \neq \lim_{n \rightarrow \infty} f(y_n)$.
Let me give an example. Consider $$f: [0,2] \rightarrow \mathbb{R}, f(x) = \begin{cases} 0, & \text{if } x \leq 1, \\ 1, & \text{if } x > 1.\end{cases}$$
Now the sequence $\{1-\frac{1}{n}\}$ and the sequence $\{1+\frac{1}{n}\}$ both converge to $1$, but $$0=\lim_{n \rightarrow \infty} f(1-\frac{1}{n})\neq \lim_{n \rightarrow \infty} f(1+\frac{1}{n}) = 1.$$
Therefore you need to show that the limits for the different sequences coincide. And if they are not the same, then we can't give a meaningful value to $\lim_{x \rightarrow c} f(x)$, because it depends on how you approach $c$. ($c = 1$ in my example)
