A second proof of the fact that a linear functional $f$ on a normed vector space $\mathcal{X}$ is bounded iff $f^{-1}(0)$ is closed.

The hint gives you an $x\in X$, with $\|x\|=1$, and such that $x\not\in\ker f$ (because this is equivalent to $\|x+\ker f\|\ne0$). Now, since $f(x)\ne0$ we can write, for any $y\in X$, $$ y=\frac{f(y)}{f(x)}\,x+\left(y-\frac{f(y)}{f(x)}\,x\right), $$ where the term in brackets belongs to $\ker f$. So $X=\mathbb{C}x\oplus\ker f$. If we write $k=|f(x)|$ we get, for any $\lambda x+m\in\mathbb{C}x\oplus\ker f$, $$ |f(\lambda x+m)|=|f(\lambda x)|=|\lambda|\,|f(x)|=k\,|\lambda|. $$ From the hint, we also have $$ \|\lambda x+m\|=|\lambda|\,\|x+\lambda^{-1}\,m\|\geq|\lambda|\,(1-\varepsilon). $$ Joining the two estimates we get $$ |f(\lambda x+m)|=k\,|\lambda|\leq\,\frac{k}{1-\varepsilon}\,\|\lambda x + m\|. $$ So $f$ is bounded.

The linear map $f$ factor through the quotient $F: X/\ker(f)\to \mathbb R$ and it is continuous since is a linear map between finite dimensional subspaces. On the other hand, the quotient map $\pi:X/\ker(f)\to \mathbb R$ is continuous as $\ker(f)$ is closed, hence $f=F\circ \pi$ so it is a composition of continuous functions.