Since $x^2-3$ has degree 2, if it is reducible, it would be the product of 2 linear factors, namely $f(x)=(x+\sqrt3)(x-\sqrt3)$.
Suppose $\sqrt3∈ℚ(\sqrt2)⇒∃a,b∈ℚ\colon\sqrt3=a+b\sqrt2⇒3=a^2+2ab\sqrt2+2b^2$
$⇒3-a^2-2b^2=2ab\sqrt2$
If $ab≠0$ then $\sqrt2∈ℚ$, contradiction.
If $b=0,a≠0$ then $\sqrt3=a∈ℚ$, contradiction.
If $a=0,b≠0$ then $\sqrt3=b\sqrt2⇒\sqrt{\frac32}∈ℚ$, contradiction.
Let $L=ℚ(\sqrt3)$, $[L:ℚ]=2$ by $\sqrt3∉ℚ$. By (a) $[L(\sqrt2):L]=2$. By Tower law $[K:ℚ]=4$.
Goal: $∃ λ ∈ 𝔽$ such that $E=F(α + λ β)$.
To show $β∈F(α + λ β)$, let $g_λ=m_{F(α+λ β),β}$, we want to show $\deg(g_λ)=1$. It is enough to show $g_λ$ has only one root, as $g_λ$ is separable.
Consider $m_{F, α}(α+λ β-λ x)∈F(α+λ β)[x]$. Since $β$ is a root of $m_{F, α}(α+λ β-λ x)$, $$g_λ∣m_{F, α}(α+λ β-λ x)$$ Suppose for a contradiction: $g_λ$ has a second root $β' ≠ β$, then $∃α'$ such that $m_{F, α}(α')=0$ \begin{aligned} α+λ(β-β') & =α' \\ ⇒ λ & =\frac{α'-α}{β-β'} . \end{aligned} The number of $α'$ is $≤\deg m_{F,α}$ so there are finitely many $λ$ of this form.Using ${|F|}=∞$ choose $λ$ not of this form to get a contradiction.
Use induction to get that $[K: F]<∞ ⇒ K=F(θ)$ for some $θ ∈ K$.