One way to prove this is by first using the primitive element theorem to show that $K = \mathbb{Q}(\alpha)$ for some algebraic number $\alpha$, whose minimum polynomial over $\mathbb{Q}$ has degree $n = [K:\mathbb{Q}]$.
Then you can in fact describe the $n$ monomorphisms $\phi_i : K \hookrightarrow \mathbb{C}$ for $i = 1, \dots, n$ by the effect they have on $\alpha$.
In particular if the minimal polynomial of $\alpha$ over $\mathbb{Q}$, say $p_\alpha(x)$ factorices over $\mathbb{C}$ as
$ p_\alpha(x) = (x - \alpha_1)\cdots (x - \alpha_n) $
with $\alpha_1 = \alpha$, then the monomorphisms $\phi_i$ are given by $\phi_i(\alpha) = \alpha_i$.
For example if you have the extension $K = \mathbb{Q}(\sqrt{5})$ then the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ is $p(x) = x^2 - 5 = (x- \sqrt{5})(x + \sqrt{5})$. Then there are only two monomorphisms given by $\phi_1(\sqrt{5}) = \sqrt{5}$ and $\phi_2(\sqrt{5}) = -\sqrt{5}$. Then since the elements of $K = \mathbb{Q}(\sqrt{5})$ are of the form $a + b\sqrt{5}$ they are given by
$ \phi_1(a + b\sqrt{5}) = a + b\sqrt{5} \quad \quad \phi_2(a + b\sqrt{5}) = a - b\sqrt{5} $
Now, if you already now the description of the monomorphisms, the proof that indeed these are monomorphisms and that every monomorphism $K \hookrightarrow \mathbb{C}$ is of this form is not difficult and you should try to complete it on your own.