Why can we say "Let $(x_n)$ be a Cauchy sequence in $X$"? I am very new to analysis. To show a space $X$ is complete, they always begin with, such as, "Let $(x_n)$ be a Cauchy sequence in $X$". But I am confused that why we can must find a Cauchy sequence in $X$, without knowing any property of $X$. In other words, there must exist Cauchy sequence in any space $X$, why? 
I know Cauchy sequences are those whose elements become closer and closer in terms of a norm, as the label goes larger and larger. While I think nothing guarantees such a sequence exists in arbitrary space. 
For example from a textbook: (Please pay attention to my remarks) 
To show a linear subspace $M$ of a Banach space $(X, \|\cdot\|)$ is complete if and only if it is closed. 
Complete $\Rightarrow$ Closed: 


*

*Let $x \in \overline{M}$, then there is a sequence $(x_n)$ in $M$ such that $\|x_n - x\| \to 0$ as $n \to \infty$. (Remark: Why is there such a sequence? Why does it guarantee that we can find such a sequence?) 

*Since $(x_n)$ converges, it is Cauchy. (Remark: This is true if $M$ is a normed linear space. A subspace of a normed space must be normed with the same norm?) 

*Completeness of $M$ guarantees the existence of an element $y \in M$ such that $\|x_n - y\| \to 0$ as $n \to \infty$. 

*By uniqueness of limits, $x = y$. 

*Hence $x \in M$ and consequently $M$ is closed. (Remark: Which one is the definition of closed set: Its complement set is open, or, the limit points of all its sequences are contained in it? If the later, $(x_n)$ is "any" sequence rather than "a" sequence in step 1, and consequently $x$ here is every limit point?) 


Complete $\Leftarrow$ Closed: 


*

*Let $(x_n)$ be a Cauchy sequence in $M$. (Remark: Why does such a Cauchy sequence exist and can be found? I mean, who can guarantee this? Indeed, here it means "a" Cauchy sequence or "any" Cauchy sequence?) 

*Then $(x_n)$ is a Cauchy sequence in $X$. 

*Since $X$ is Banach and thus complete, there is an element $x \in X$ such that $\|x_n - x\| \to 0$ as $n \to \infty$. 

*But then $x \in M$ since $M$ is closed. (Remark: Is this by definition of closed set I mentioned?)

*Hence $M$ is complete. (Remark: It seems not "a" but "every" Cauchy sequence in step 1?) 


As I said, I am very new to analysis. I am really hoping someone can help me to review and answer my remarks carefully. Thank you in advance for your patience. 
 A: In any non-empty space $X$ there are Cauchy sequences. Just take some $x\in X$ and define $(\forall n\in\mathbb{N}):x_n=x$.
Concerning the question about why $x\in\overline M$ implies that there is a sequence $(x_n)_{n\in\mathbb N}$ that converges to $x$, that's because that's one of the properties of the closure (some authors take it as the definition of closure).
And, yes, when we talk about a subspace of a normed space, the default assumption is that the norm is the same as that of the original space.
Finally, concerning the definition of closed set, it's up to you to tell us which definition that textbook of yours contains. It's not as if all textbooks use the same definition.
A: If there are no Cauchy sequences, the space is automatically complete, so there is nothing to prove.
A: You must show that:
$(x_n)$ cauchy sequence in $X \Rightarrow (x_n)$ converges
The implication $p \Rightarrow q$ is true if $p$ is false, so if there are no cauchy sequences, the statement is trivially true.
Of course, in a non empty space $X$, any constant sequence is a cauchy sequence, so you can always find a cauchy sequence.
A: Part of what's going on here is illustrated by this situation: An instructor asks whether all cell phones in the classroom are turned off. If it happens that there are no cell phones in the classroom, then the answer is "yes". That makes sense, as you may see if you think about it.
Does every Cauchy sequence in $X$ converge? To answer "yes" to that question, means that if there is at least one such Cauchy sequence then it, and all others, converge. Just as "yes" in the previous paragraph means that if there is at least one cell phone in the classroom, then it, and all others, are turned off.
However, it is also the case that every non-empty metric space contains at least one Cauchy sequence, since every constant sequence is a Cauchy sequence.
