Proof that a compact metric space is sequentially compact I have tried to construct a proof that is different from the others given on this site. It would be great if you could tell me whether the reasoning is sound.
Let $\langle x_{n}\rangle$ be a non-convergent sequence of points. Let the set $\{1,a,b,c,d,\dots\}\in\Bbb{N}$ be strictly increasing. 
For $x_{1}$, there is an $\epsilon_{1}\in\Bbb{R}$ such that $B(x_{1},\epsilon_{1})$ does not contain all the points in $\langle x_{n}\rangle\setminus x_{1}$. This is because if for all real numbers $r$ the ball $B(x_{1},r)$ contained every other element in the sequence, then the sequence would be convergent.
Let $x_{a}$ be a point in $\langle x_{n}\rangle\setminus B(x_{1},\epsilon_{1})$. Then there is an $\epsilon_{2}\in\Bbb{R}$ such that $B(x_{i},\epsilon_{2})$ does not contain all the points $x_{j}$ where $j>a$. More importanty, we can ensure that $\epsilon_{2}<\epsilon_{1}$ 
Let $x_{b}$ be a point in $\langle x_{n}\rangle\setminus B(x_{a},\epsilon_{2})$. Proceeding like this, we can find a non-finite cover of $X$ which does not have a finite subcover, as each set contains a point ($x_{i},i\in\{1,a,b,c,\dots\}$) that no other does.
Is this argument sound? 
Thanks for your time!
 A: The basic idea of your argument is ok, except that (as pointed out by Ittay) you are not producing a cover. However, if you add the complement of the sequence $(x_n)$ to your cover, then you are fine. Note that this is an open set because the sequence has no convergent subsequence by assumption.
A somewhat simpler argument would be to consider the open sets
$$U_k := X\setminus\{x_j:j\geq k\}.$$
These form a cover of your space. Since the sequence must be infinite, none of these sets equals the whole space, and hence there is no finite subcover.
Of course this boils down to the same idea as any other proof. 
A: Firstly, instead of considering the set $\{1,a,b,c,d,\cdots \}$, which then forces you to define $x_a,x_b$ etc. just consider the set $\mathbb N$, and then you simply define $x_2,x_3$ and so on. 
Regardless, it is not clear what is the cover of $X$ that your argument produces.
I must add that I don't know which other proofs you are referring to, but the proof I have in mind is very short. 
