$X$ is a Banach space and $f$ be a linear functional from $X$ to $\mathbb{C}$. If $\operatorname{Ker} f$ is a $G_\delta$ set in $X$, then $f$ is continuous.
I know that if $\operatorname{Ker} f$ is closed, then $f$ is continuous. The argument is based on getting a ball of radius $r$ in $(\operatorname{Ker} f)^c$ and then showing that $|f(x)|\le \frac1r||x||$.
I was trying to extended this idea to the above problem, but failed. Any hint/ suggestions.
Suppose $f$ is unbounded. Then its kernel is dense in $X$ (as noted by Mechanodroid). Let $U = X\setminus\operatorname{Ker} f$; by assumption this is a countable union of closed sets $F_n$, $n\in\mathbb N$. Each $F_n$ has empty interior, because $\operatorname{Ker}f $ is dense.
Pick any vector $x_0$ such that $f(x_0)\ne 0$. Note that every point $x\in X$ is either in $U$ or in $U+x_0$. Hence, $$ X = \left(\bigcup_{n=1}^\infty F_n \right) \cup \left(\bigcup_{n=1}^\infty (F_n +x_0)\right) $$ But a complete metric space cannot be written as a countable union of closed sets with empty interior, by Baire category theorem.