I am looking to prove that the category of vector spaces is not autodual, i.e., equivalent to its opposite category, in the simplest way. Here are my ideas.
We have an equivalence between finite-dimensional $k$-vector spaces and its opposite category, given by its dual construction
$$ h_k\colon \mathbf{fVec}_k\to\mathbf{fVec}^{op}_k, X\mapsto \mathbf{Vec}_k(X,k). $$
Then since equivalences are essentially unique, if $F\colon \mathbf{Vec}_k \to \mathbf{Vec}_k^{op}$ is an antiequivalence, it must restrict to the dual on finite-dimensional objects.
We also make the observation that for $F$ to be an equivalence it should restrict to an isomorphism of skeletons (a category where all isomorphic objects are identified). So it is sufficient to show for arbitrary powers of $k$.
We check where $F$ takes $\bigoplus_{i\in \mathbb N} k$. Since $F$ is assumed to be an antiequivalence, it must take colimits to limits. Therefore $F\left(\bigoplus_{i\in \mathbb N} k \right)=\prod_{i\in\mathbb N}Fk$. But $Fk=\mathbf{Vec}_k(k,k)\cong k$. Thus, since $\bigoplus_{i\in \mathbb N}k\ncong \prod_{i\in\mathbb N}k$, $F$ cannot be an antiequivalence.
Is this the easiest approach, or is there another categorical property that one could use to show that the category of vector spaces is not autodual?