I think I figured it out. The proof the set $\Omega$ is convex is very easy. Now the next part, let $x^* \in H$ then $x^*=\Sigma(x^*,e_k)e_k$ and so $(x^*,x)=\lim_{n\to \infty}(\sum_{0}^{n}(x^*,e_k)e_k,\sum_{0}^{m}c_ke_k)$ Now we can assume $n>m$ and make use of linearity and orthonormality of $e_k$ to conclude that $(x^*,x)=\lim_{n\to \infty}(\sum_{0}^{n}(x^*,e_k)e_k,\sum_{0}^{m}c_ke_k)=\sum_{0}^{m}(x^*,e_k)c_i$ Now we know this sum is not $0$ for any choice of $m$ and $c_i$ as the would imply $x^*=0$ thus we have at least one is non zero. Thus we can use that non-zero part, to get any real number (it is a little more nuanced but i am sure i got this part right).
Is this correct?
There are notational incosistencies in your argument and the logic itslef is not clear. Where did you use the fact that last coffecient is non-zero?
Here is proof: Suppose $x^{*}(e_k) \neq 0$. If $x^{*}(e_k) > 0$ then $x^{*} (-ne_k+e_{k+1})\to -\infty$ and $x^{*} (ne_k+e_{k+1})\to \infty$. Now $x^{*}(\Omega)$ is a convex set in $\mathbb R$ so it is an interval. Hence it contains all real numbers. Similar argument holds when $x^{*}(e_k) <0$.