Topological Vector Space and $\phi : X \to \mathbb{K}$ linear form such that $\phi$ is not continuous The following question was asked in my assignment of Functional Analysis course and unfortunately no question in  assignment was discussed by the professor.

Question: Let $X$ be a topological vector space over $\mathbb{K} = \mathbb{C}$ or $ \mathbb{R}$ and let $\phi: X\to \mathbb{K}$  be a linear form. Show that if $\phi$ is not continuous , then $\phi (W)  $ equals $\mathbb{K} $ for any neighbourhood W of 0 in X and deduce that $ker(\phi)$ is dense in $X$.

Let $\phi$ be not continuous. I am very sorry but I have no idea on how to prove the statement asked in the question. I am not able to find any proposition in class notes which I can use here. I couldn't find much of use in this particular question by googling.
I really don't know which result should be used. Please don't vote to close this question! Please guide me!
 A: As pointed out in the comments, if $\phi$ is not continuous, then it is not continuous at $0$.
The following is a standard fact about topological vector spaces, and it is proven using the basic properties of its topology:
The origin of a topological vector space admits a basis of balanced neighbourhoods.
*We say that a set $X$ is balanced if $\alpha X\subset X$ for every $|\alpha|\leq 1$.
Now, take $W$ a neighbourhood of the origin. We can assume that it is balanced. Let $k\in \mathbb{K}$, $k\neq 0$.  Since $\phi$ is not continuous at $0$, $\phi(W)$ cannot be bounded (why?!), so we can find $x\in W$ such that $|\phi(x)|\geq |k|$. As $W$ is balanced, $(k/\phi(x))x\in W$, and $$ \phi\left( \frac{k}{\phi(x)}x\right)= k$$
So we get $\phi(W)=\mathbb{K}$, as wanted.
The second part is an immediate consequence of the first. For if we take any $x\in X$ and any neighbourhood $W$ of $x$, $W-x$ is a neighbourhood of $0$, and by what we have just proven, $\mathbb{K}=\phi(W-x)=\phi(x)+\phi(W-x)=\phi(W)$.
In particular, there is $y\in W$ with $\phi(y)=0$. This proves that $\operatorname{ker}(\phi)$ is dense in $X$.
A: You might start out by showing that if $\phi$ is discontinuous, it is discontinuous at $0$.  Next, note that for any $s, t \in \mathbb K$ with $s \ne 0$,
$W$ contains a point $x$ with $\phi(x) = t$ iff $sW$ contains $sx$ with $\phi(sx) = st$.
