There is another post with this question which ask for a proof not using embeddings: If $\alpha$ separable over $F$ then $F(\alpha )/F$ is a separable extension..
I would like just to know the traditional proof using embeddings, as my lecture notes in my fields and Galois theory course left this reasoning for the reader.
My reasoning is the following:
With embeddings we know that $K(\alpha)/K$ is separable if and only if $|Hom(K(\alpha)/K,\bar{K}/K)|=[K(\alpha):K]$, and this is guaranteed cause $Irr(\alpha,K)$ has $n=[K(\alpha):K]$ distinct roots thanks to separability, let them be $\{ \alpha_i \}_{i=1}^{n}$ and for each we can construct a unique embedding such that $\alpha\longmapsto \alpha_i$, so there is exactly $[K(\alpha):K]$ of these embeddings.
Is there any nuance I should also consider?