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I conjectured that: if $E$ is a reflexive Banach space and $F: E \to \mathbb{R}$ a convex Gateaux differentiable map (in other words all the directional derivatives $\frac{\partial F}{\partial \xi}(u)$ exists continuous in $u$ and linear in $\xi$) then $F$ has a minimum on weak compact sets of $E$. I found a proof but I need a confirmation.

Let $K \subseteq E$ be a weak compact sets of $E$. Take $u_n \in K$ such that $F(u_n) \to \inf_{x\in K} F(x)$. Because of the reflexivity of $E$ and the weak compactness of $K$ you can assume that $u_n \to u \in K$ weakly. I claim that $F$ attain its minimum at $u\in K$. Because the convexity of $F$ we have that $F(u_n) \geq F(u)+ \frac{\partial F}{\partial (u-u_n)}(u)$. Taking the limit for $n \to \infty$ in this inequality we obtain that $\inf_{x\in K} F(x) \geq F(u)$, so $F(u)= \inf_{x\in K} F(x)$ and we are done.

Antonio Alfieri
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Your proof is correct. Here is another one, which does not need the differentiability, but needs lower semicontinuity of $F$. This is a weaker assumption, since the inequality $F(u_n) \geq F(u)+ \frac{\partial F}{\partial (u-u_n)}(u)$ implies lower semicontinuity at $u$.

Proceed as you did until $u_n\to u\in K$. Then use Mazur's lemma to obtain a sequence $v_n$ that converges strongly to $u$, and consists of convex combinations of $u_m$, $m\ge n$. (Note that we don't know if $v_n\in K$, but this does not matter.) By convexity, $F(v_n)\le \sup_{m\ge n} F(u_m)$. Hence, $$\liminf F(v_n)\le \lim F(u_n) = \inf_K F$$ By lower semicontinuity of $F$ we conclude that $F(u)\le \inf_K F$, hence the infimum is attained.