In Corollary 7.1.5 in Voight's Quaternion Algebras, the author gives a direct proof of Skolem-Noether in the case of quaternion algebras. The only challenging step is showing that for a division quaternion algebra $B$ over a field $F$, if $\alpha_1$ and $\alpha_2$ have the same minimal polynomial, then they are conjugate. To do this, the author defines the $F$-vector space $$W = \{\beta \in B : \beta \alpha_2 = \alpha_1 \beta\}.$$ He then shows that $\alpha_1$ and $\alpha_2$ are conjugate in $B \otimes_F F^{\mathrm{sep}}$, and concludes:
but then by linear algebra $\operatorname{dim}_{F^{\mathrm{sep}}} W \otimes_F F^{\mathrm{sep}} = \operatorname{dim}_F W > 0$, so there exists $\beta \in B \setminus \{0\} = B^\times$ with the desired property.
I see why $\operatorname{dim}_{F^{\mathrm{sep}}} W \otimes_F F^{\mathrm{sep}} = \operatorname{dim}_F W$ by properties of tensor products, but I'm not sure how to use this to conclude that $\operatorname{dim}_F W > 0$ (which is what we are trying to prove, since this implies there is $\beta \in B \setminus \{0\}$ with $\beta \alpha_1 = \alpha_2 \beta$). In particular, if $W$ were $\{0\}$, which is what we are trying to disprove, then wouldn't we have that $\operatorname{dim}_F W = 0$?