Let $E$ be an infinite-dimensional t.v.s. and $E^\star$ its topological dual. Let $B$ be a basis of $E$, i.e., each $x\in E$ is a finite linear combination of some elements in $B$ and each finite subset of $B$ is linearly independent.
For each $x \in E$, there is a unique collection $(\lambda_{x,b})_{b\in B}$ such that $x = \sum_{b\in B}\lambda_{x,b} b$. By definition, $(\lambda_{x,b})_{b\in B}$ has finitely many non-zero terms. For each $b \in B$, we define $f_b:E \to \mathbb R$ by $f_b(x) := \lambda_{x,b}$. Clearly, $f_b$ is linear. We have
$$f (x) = f \left ( \sum_{b\in B}\lambda_{x,b} b \right ) = \sum_{b\in B}\lambda_{x,b} f \left ( b \right ) = \sum_{b\in B} f(b) f_b (x), \quad \forall f \in E^\star, x\in E.$$
If $\{f_b \mid b\in B\} \subseteq E^\star$, then $\dim E \le \dim E^\star$ would be true. However, If $E$ is a Banach space, then there are only finitely many $b$'s in $B$ such that $f_b$ is continuous.
Could you elaborate on this point?