I am working with the weak topology on $\ell^2(\mathbb{N})$. I have a set $K$ such that for every linear functional $\phi \in \ell^2(\mathbb{N})^*$, there exists a sequence $\{v_n\}$ in $K$ so that $\phi(v_n) \rightarrow 0 $. I want to show that every neighborhood $U$ of $0 \in \ell^2$ in the weak topology intersects $K$, i.e. $K \cap U \neq \emptyset$.
My attempt so far is to look at the basis neighborhoods of $0$, which are defined for me as $$\mathcal{N}_{0, \phi_1, \ldots, \phi_k, \epsilon} = \{v \in \ell^2: |\phi_i(v)| < \epsilon \text{ for all $i = 1, \ldots, k$}\}.$$ I've been attempting a contradiction, maybe by showing that there is a $\phi_i$ such that $\phi_i$ does not converge to $0$ for any sequence in $K$ if we assume $K \cap U$ is empty. But I haven't had any success.
For additional context, $K$ is defined to be $K = \bigcup_{k=2}^\infty \{\log(k)e_j\}_{j=1}^k$ where $e_j \in \ell^2$ is the vector with $1$ in the jth position and $0$ everywhere else. So the solution might be dependent on how $K$ is defined.