TITLE: How to finish this proof about compact implies bounded
QUESTION [5 upvotes]: A set is called compact if every sequence has a convergent subsequence. I am trying to show: If $K$ is compact then it is bounded. (that it is closed was very easy to prove)
What I want to do:
Let $K$ be compact and assume that it is  not bounded. Let $x \in K$ be any point. Then for every $n \in \mathbb N$ let $x_n$ be such that $d(x,x_n) > n$. Now I want to show that $x_n$ does not have a convergent subsequence. I tried by contradiction: assume $x_{n_k}$ was a convergent subsequence and $x_{n_k}\to y$ for some $y \in K$. 
How to proceed?
Edit: please note that I know how to prove it using the cover defintion and I am not asking for a proof using the equivalent defintion.

REPLY [2 votes]: To keep in touch with your previous work, 
if you suppose that $x_{n_k}\to y$, note that the function $h\to d(x,h)$ is continuous.
Therefore, the sequence $d(x,x_{n_k})_k$ converges to $d(x,y)$.
But since $d(x,x_{n_k}) > n_k$ and the sequence $(n_k)_k$ is increasing, the sequence $d(x,x_{n_k})$ is unbounded and convergent.
Contradiction.

A different approach
Suppose for contradiction that $K$ is unbounded.
Negating the definition of boundedness, one derives the existence of two sequences $(x_n)$, $(y_n)$ such that $\forall n\in \mathbb N, d(x_n,y_n)>n$.
Since $K$ is compact, there is a convergent subsequence for $(x_n)$, say $\large(x_{n_k})$.
The first trick: since $K$ is compact, the sequence $\large(y_{n_k})$ has a convergent subsequence, say $\large(y_{n'_j})$ (I changed notations for clarity).
The second trick: the sequence $\large(x_{n'_j})$ is convergent since it is a subsequence of $\large (x_{n_k})$.
Therefore, the sequence $(\large d(x_{n'_j},y_{n'_j}))$ converges, and is unbounded.
Contradiction.