Uniform continuity and sequentially compact space Let $(X,d)$ be a sequentially compact metric space and let $(Y,d')$ be a metric space. Let $f: X \to Y$ be a continuous function. Then,$f$ is uniformly continuous.
Just to be sure, by sequentially compact I mean: From any sequence in $X$, there is a convergent subsequence.
I feel this is so simple, but still I'm missing something. We let $f$ be indeed continuous, and we assume $f$ is not uniformly continous. By the proof, it just looks like we are showing "continuity implies uniform continuity".
$f$ is not uniformly continuous: 
$\exists \epsilon \gt 0$ : $\forall n \gt 0, \exists x_{n}, y_{n} \in X, d(x_{n},y_{n}) \lt \frac{1}{n}, d'(f(x_{n}),f(y_{n})) \gt \epsilon$.
Then, we've built in the mean time two sequences $(x_{n})_{n}$ and $(y_{n})_{n}$
, converging both to say $l$.
Yet, $d'(f(x_{n}),f(y_{n})) \gt \epsilon, \forall n \in \mathbb N $, which contradicts the continuity of $f$.
How is it important to know $(X,d)$ is sequentially compact ? It seems to me we are not using this assumption at all. We just take the "discrete" definition of uniform continuity to get two convergent sequences leading us to the contradiction. So this would show continuity is anytime equivalent to uniform continuity (of course not).
Any help appreciated.
 A: You made an unjustified jump when you asserted that the sequences $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ converge. Why?
Take a subsequence of $(x_n)_{n\in\mathbb N}$ which converges to some $l$. Then the similar subsequence of $yx_n)_{n\in\mathbb N}$ will converge to $l$ too. What I mean by “similar” here is this: if your sebsequence of $(x_n)_{n\in\mathbb N}$ is $(x_{n_k})_{k\in\mathbb N}$, take the sequence $(y_{n_k})_{k\in\mathbb N}$.
The rest is more or less right. You should justify why there is a contradiction with the continuity of $f$. And when you deny that $f$ is uniformly continuous, you should have written $\geqslant\epsilon$ instead of $>\epsilon$.
A: You need the sequential compactness. The fact that $d(x_n, y_n) < \frac1n$ doesn't say anything about covergence of the sequences. It just means they are “equiconvergent” – if one converges to $l$, than the other converges to $l$ as well. You have to use sequential compactness to move to a convergent sumsequences $x_{n_k}$, $y_{n_k}$.
A: When you build these two sequences $x_n$ and $y_n$ you will need the compactness of $X$. Otherwise they do not have to converge. Think for example of $x_n=n$ and $y_n=n+\frac{1}{n}$
