It is clearly enough to show that an infinite dimensional vector space $V$ has smaller dimension that its dual $V^\*$.

Let $B$ be a basis of $V$, let $\mathcal P(B)$ be the set of its subsets, and for each $A\in\mathcal P(B)$ let $\chi_A\in V^\*$ be the unique functional on $V$ such that the restriction $\chi_A|_B$ is the characteristic function of $A$. This gives us a map $\chi:A\in\mathcal P(B)\mapsto\chi_A\in V^\*$.

Now a complete infinite boolean algebra $\mathcal B$ contains an independent subset $X$ such that $|X|=|\mathcal B|$---here, that $X$ be independent means that whenever $n,m\geq0$ and $x_1,\dots,x_n,y_1,\dots,y_m\in X$ we have $x_1\cdots x_n\overline y_1\cdots\overline y_n\neq0$.

If $X$ is such an independent subset of $\mathcal P(X)$ (which is a complete infinite boolean algebra), then $\chi(X)$ is a linearly independent subset of $V^\*$, as one can easily check. It follows that the dimension of $V^*$ is at least $|X|=|\mathcal P(B)|$, which is strictly larger than $|B|$.