Compactness / sequentially compact / reflexive space / separable spaces / weak topology I'm a bit confuse with all theses notions. Let $E$ a normed vector space of infinite dimension (also Banach, but it's probably not important).

The theorem of Eberlin Smulian theorem says that : all bounded sequence that has a subsequence that converge weakly $\iff$ it's reflexive. 

(In fact it just says implication, but the converse is also true)... anyway.
Q1) Does it mean that if $E$ is reflexive, then instead of the fact that the weak topology is not metrizable, the property 

$C\subset E$ is compact $\iff$ $C$ is sequentially compact 

hold ? Because in reflexive spaces, $\{x\in E\mid \|x\|\leq 1\}$ is compact. Eberlin Smulian theorem says that  

$\{x\in E\mid \|x\|\leq 1\}$ is compact $\iff$ it's sequentially compact.

Can this be generalized for any compact $C$ ?
Q2) If a set is separable, we know that $\{x\in E\mid \|x\|\leq 1\}$ is metrizable for the weak topology. In particular, can we conclude from this that 

If $E$ is separable, a set $C\subset E$ is compact $\iff$ it's sequentially compact.

 A: Eberlein -Smulian is usually stated in the form:

Let $A$ be a subset of a Banach space $X$. Then, $A$ is weakly compact
  if and only if $A$ is weakly sequentially compact.

The form you gave is a consequence of this, as the unit ball is weakly compact if and only if the space is reflexive. But, since the form you gave says absolutely nothing about non-reflexive spaces, it is actually weaker than E-S.
Q1) The answer is true in general, by the (stronger) ES, and it does not require reflexivity.
Note that, if I remember right, the key for the proof is the following:
If $E$ is a separable Banach space, then $E*$ contains a countable total set. In this case, the weak topology becomes metrisable on weakly compact sets.
Now if $x_n$ is any sequence, the subspace spanned by $\{ x_n \}$ is separable, and the trick above shows that weak-compactness implies weak sequential compactness.
The other implication is a bit trickier, it usually is done by showing that if $A$ is weakly sequentially compact it is bounded and the weak* closure of $A$ is included in $E$. 
