A finite Galois extension is a finite, separable, normal extension. Any subextension of a separable extension is separable, similarly for finite extensions (mostly obviously). However, it is certainly not true for normality.
Consider arbitrary $F$, $K$ the splitting field of a minimal (separable) polynomial in $F[X]$ with at least two distinct roots. Then $F[\alpha]$, where $\alpha$ is just one of the roots of the polynomial is not normal.
But if you want $E$ to be generated by some roots of some polynomial, then of course this is true. You don't even need $K$ to be Galois, just finite (and hence algebraic). Any finite field extension is algebraic, so for arbitrary $F\subseteq E\subseteq K$ we have $E=F[\alpha_1,\ldots,\alpha_n]$ (this is true for example because $E$ is a subspace of $K$ as a vector field over $F$), an then $\alpha_j$ are algebraic, so if you take for $p$ the product of their respective minimal polynomials over $F$, they will all be roots of $p$.
Furthermore, if $K$ was separable, then if you take only distinct minimal polynomials of $\alpha_j$ (so you won't take $x^2+1$ twice for $\alpha_1=i,\alpha_2=-i$, for instance), $p$ will also be separable (because any $\alpha_j\in K$ is separable).