Let $E$ be topological vector space on field $\mathbb{R}$(or $\mathbb{C}$), which need not be Hausdoff. $f$ is a linear functional on $E$, and there are open set $U\subset E$ and $t\in \mathbb{R}$(or $\mathbb{C}$) ,such that $t\notin f(U)$.
How to prove $f$ is continous?
By translation, you can assume $0\in U$ and, using the continuity of the multiplication $\mathbb K \times E\to E$ ($\mathbb K =\mathbb R$ or $\mathbb C$), you get a symmetric $0$-neighborhood $V\subseteq U$ (where symmetry means $\alpha v\in V$ for all $v\in V$ and $|\alpha|\le 1$). Clearly $t\notin f(V)$ and the symmetry then implies $|f(v)|\le |t|$ for all $v\in V$ which gives the continuity.