Assume that we have a sequence of complex numbers $\alpha=(\alpha_k)_k\in\ell^\infty$.
Since $(\ell^2)^*=\ell^2$, we know that $$\sup_{\|\beta\|_2=1}=|\langle\alpha,\beta\rangle|=\|\alpha\|_2,$$ where $\|\alpha\|_2=(\sum_k|\alpha_k|^2)^{1/2}$ and $\langle{\alpha,\beta}\rangle=\sum_k\alpha_k\overline{\beta_k}$.
The question is the following: if we have that $$\displaystyle\sup_{\substack{\|\beta\|_2=1\\\beta\in c_{00}}}=|\langle\alpha,\beta\rangle|=C<\infty,$$
where $c_{00}$ is the space of complex sequences with finite support, can we say that $C=\|\alpha\|_2$? Do we need any assumption on $\alpha$ for it?
Possible answer:
Clearly, $C\leq\|\alpha\|_2$ since $c_{00}\subset\ell^2$.
On the other hand, to see that $\|\alpha\|_2\leq C$, I think about using the density of $c_{00}$ in $\ell^2$ and triangle inequality as follows:
By density, for each $\beta\in\ell^2$ we can find $\gamma_\beta\in c_{00}$ such that $\|\beta-\gamma_{\beta}\|_2<\varepsilon$. Now,
$$|\langle \alpha,\beta\rangle|\leq|\langle \alpha,\beta-\gamma_\beta\rangle| + |\langle \alpha,\gamma_\beta\rangle|.$$
but to ensure that the first term on the rhs is small and take supremums on beta after, don't you have to assume that $\alpha\in\ell^2$ from the start?
We try to circunvent that by taking the truncated sequences $\alpha^{k)}:=(\alpha_j)_{j=1}^k$. Now
$$|\langle \alpha^{k)},\beta\rangle|\leq|\langle \alpha^{k)},\beta-\gamma_\beta\rangle| + |\langle \alpha^{k)},\gamma_\beta\rangle|\leq \| \alpha^{k)}\|_2\|\beta-\gamma_\beta\|_2 + |\langle \alpha^{k)},\gamma_\beta\rangle|,$$
Therefore
$$|\langle \alpha^{k)},\beta\rangle|-\varepsilon\cdot\| \alpha^{k)}\|_2\leq |\langle \alpha^{k)},\gamma_\beta\rangle|.$$
Now taking limits on k and supremums on $\beta$, we conclude that
$$(1-\varepsilon)\|\alpha\|_2\leq C,$$
so $\|\alpha\|_2\leq C$.
The norm of continuous linear operator can always be estimated on a dense subspace. Since $c_{00}$ is a dense subspace of $\ell^2$ it follows that $\sup\limits_{||\beta||_2=1, \text{ }\beta\in c_{00}}|\langle \alpha, \beta \rangle|=\|\alpha\|_2$.