At most finitely many (Hamel) coordinate functionals are continuous - different proof

How about this: If you have a Hamel basis $\{x_i\}$, and replace each $x_i$ by a nonzero scalar multiple of itself, then the result is still a Hamel basis. The corresponding functionals $a_i$ are of course replaced by nonzero scalar multiples of themselves (the multiplier for $a_i$ is the reciprocal of the multiplier for $x_i$). The new functional is continuous iff the original was continuous. If $J$ is infinite, then you can carry out this "replacement" by scalar multiples in such a way that the $\sup$ in (c) is infinite.

Speaking of different proofs, there is also a simple argument using Baire's Category Theorem.

Let $\{b_i:i\in I\}$ be a Hamel basis of $X$ and suppose that $(b_n^{\#})_{n\in\mathbb{N}}$ is a sequence of bounded coordinate functionals. Then $\bigcup_{n=1}^\infty \ker b_n^{\#}=X$.

This is easy to check since for each $x\in X$, $x=\sum_{i\in F}\lambda_ib_i$ for some finite $F\subseteq I$, so there exists an $n_0\in \mathbb{N}$ such that $b_{n_0}$ does not appear in this expression. Then $x\in \ker b_{n_0}^{\#}$.

Since $b_n^{\#}$'s are bounded, their kernels are closed, so by Baire's Theorem there exists an $n_0\in\mathbb{N}$ such that $\ker b_{n_0}^{\#}$ has non empty interior. This implies that $\ker b_{n_0}^{\#}=X$, which is a contradiction, since $b_{n_0}\notin \ker b_{n_0}^{\#}$.