Is there an example of a TVS on which there is no nonzero continuous linear functional.

Let $f: X \rightarrow K$ be a nonzero continuous linear functional on a topological vector space (TVS).

Prove that the set $U = \{x \in X: f(x) < 1\}$ is an open convex set in $X$ containing 0.

Bonus: Give an example of a TVS on which there is no nonzero continuous linear functional. (Hint: It should contain no open convex neighborhood of 0. Try $\mathbb{K}^2$ with the topology defined by the metric

$$d((x_1, x_2), (y_1, y_2)) = (\sqrt{|x_1 - y_1|} + \sqrt{|x_2 - y_2|})^2.$$)
I have proved the first question, but i dont know how to give a example according to the hint, the question want me to give the example by using the property I just proved, maybe some contradiction exsit.