Let $X$ a Banach space. Let $\mathcal T$ the weak topology on $X$ (the thickest topology s.t. linear form are continuous).
(1) $A\subset X$ is weakly compact if for all covering $\mathcal U\subset \mathcal T$ there are $U_1,...,U_n\in \mathcal U$ s.t. $$A\subset \bigcup_{i=1}^n U_i.$$
(2) $A\subset X$ is sequentially weakly continuous if for all $(x_n)\subset A$ there is a subsequence that converge weakly, i.e. there is $(x_{n_k})\subset (x_n)$ and $x\in A$ s.t. for all $\varphi\in X^*$ (where $X^*$ denote the topological dual) $$\lim_{k\to \infty }\varphi(x_{n_k})=\varphi(x).$$
Are (1) and (2) equivalent ?