I'm following the proof in http://www.science.oregonstate.edu/~parks/sample338.pdf .
Theorem 5 If V is infinite dimensional, then V and $V^∗$ are not isomorphic.
Proof $B$ basis of $V$, $\kappa=|B|$. Case I $2^{\aleph_0}\leq \kappa$. For each subset $S\subset B$ we can define a distinct element $$ \sum_{b\in S}b^*. $$ Thus we see that the cardinality of $V^∗$ is at least $2^\kappa$ which is strictly greater than $\kappa = \kappa\cdot 2^{\aleph_0}$, the cardinality of V. Case II not relevant for the question.
What is the relation between $|B|$ and $|V|$?
I understand that $|B|=\kappa\geq 2^{\aleph_0}$ is given, and that $|V^*|\geq 2^\kappa$. Additionally I have the identity for infinite $V$ $$ |V|=\textrm{max}(|\mathbb{F}|,|B|), $$ where $\mathbb{F}$ is the field under the vector space. Am I correct if I say that the author makes the assumption $|\mathbb{F}|\leq 2^{\aleph_0}$?
And another question (to understand where $\kappa = \kappa\cdot 2^{\aleph_0}$ comes from): Is $|V|=\textrm{max}(|\mathbb{F}|,|B|)$ equivalent to $|V|=|\mathbb{F}|\cdot|B|)$?
Twice yes: the first line of the linked note says we are considering vector spaces over the real numbers. So $|\mathbb{F}| = 2^{\aleph_0} = \mathfrak{c}$
And for all infinite cardinals $\kappa, \lambda$ $$\kappa \cdot \lambda = \kappa + \lambda = \max(\kappa, \lambda)$$
(at least if we assume the axiom of choice (AC), which we do, because we assume a vector space always has a basis)
So for all vector spaces over an infinite field $|V| = |\mathbb{F}| \cdot |B|$ (because we use finite linear combinations from $B$ to get all of $V$), so over the reals : $|V| = \mathfrak{c} \cdot |B|$ and so in case (i), when $|B| \ge 2^{\aleph_0}$: $|V| = |B|$, while the argument shows the first inequality in $$|V^\ast| \ge 2^{|B|} > |B| = |V|$$ the last inequality by Cantor's theorem.