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 $E$.
The result is true and your argument is correct. A slightly better phrasing might be to show that every subsequence of $u_n$ has a further subsequence that converges weakly to $u$ in $H^1$. It is then a standard topological result that this implies that $u_n \rightharpoonup u$. In fact, the last part of your argument would essentially prove the general topological result.
Perhaps the reason that you cannot find this exact result anywhere is that it is a not too difficult instance of the well known criterion that says that a relatively compact sequence with a unique limit point must converge to that limit point.
Best Answer
The property you are looking for is true in every reflexive ordered Banach space $V$ (provided that the cone $V_+$ is closed).
Proof. Let $(v_n)$ be an increasing norm bounded sequence in $V$.
Let us first note that if $(v_{n_k})$ and $(v_{m_k})$ are subsequences of $(v_n)$ and if they converge weakly to vectors $v$ and $w$, respectively, then $v = w$.
Indeed, let $v' \in V'$ be a positive and continuous linear functional on $V$. Both real-valued sequences $$ \big( \langle v', v_{n_k} \rangle \big) \quad \text{and} \quad \big( \langle v', v_{m_k} \rangle \big) $$ are subsequences of the increasing sequence $(\langle v', v_n \rangle)$, so their limits $\langle v', v \rangle$ and $\langle v', w \rangle$ coincides. Since the span of the dual cone is weak${}*$-dense in the dual space $V'$, we conclude that even $\langle w', v \rangle = \langle w', w \rangle$ for every functional $w' \in V'$, be it positive or not. Thus, $v = w$.
Now the claim easily follows: By the reflexivity, every subsequence of $(v_n)$ has a weakly convergent subsequence, and all these weakly convergent subsequences of subsequences have the same weak limit - say, $v$ - by what we have shown above. Thus, the sequence $(v_n)$ itself converges to $v$. $\square$
Remarks.
(1) I assumed that by a bounded sequence you mean a norm bounded sequence. If you mean order bounded instead, the situation changes considerably.
(2) The result and proof above do not assume that the ordered Banach space $V$ is lattice ordered.