I have to prove that $\text U(n)$ is a subgroup of the group $\text{SO}(2n)$.
What I have tried is just splitting a vector $x\in\mathbb C^n$ in a vector in $\mathbb R^{2n}$ by just separating the imaginary and real parts of each component of $x$, let's say $\phi:\mathbb C^n\rightarrow \mathbb R^{2n}$ is a function that identifies that (and it's clearly bijective), for example, if $y=\phi(x)$, then:
$$\Re (x_i)\mapsto y_{2i-1},$$ $$\Im (x_i)\mapsto y_{2i}.$$
From this, I am able to prove that for every $U\in\text U(n)$ there must be a linear operator $A\in\text{GL}_{2n}(\mathbb R)$ such that:
$$\phi(Ux)=A\phi(x).$$
That's simple, and it's also simple to prove that $A$ must satisfy that $A^\text TA=I$, but I can't prove that the determinant of $A$ must be $1$, I don't see why it can't be $-1$.
Prove that the map $U\to A$ is continuous, and then use the fact that $U(n)$ is connected to deduce that it's image must be one of the two connected components of $O(2n)$, and since the identity goes to the identity, the component you get is the one you want, i.e., the "determinant-equal-to-$1$" component, i.e., $SO(2n)$.