I want to show that $O(n)$ is a closed subgroup of $U(n)$. I am by confused since $U(n)$ lives in $\text{Mat}(\mathbb C,n)$ and $O(n)$ lives in $\text{Mat}(\mathbb R,n)$. It is not to hard to show that $O(n)$ is a subgroup, but I can't see why it is closed as a topological space.
To show that $O(n)$ is a subgroup of $U(n)$. There is a natural way in which $\text{Mat}(\mathbb R,n)$ lives in $\text{Mat}(\mathbb C,n)$. If $O\in O(n)$ then $OO^T=I$, but also $O^{*}=O^T$, so $O\in U(n)$.
The usual interpretation of $\mathbb R$ as a genuine subfield of $\mathbb C$ allows us to regard $\text{Mat}(\mathbb R,n)$ as a genuine subset of $\text{Mat}(\mathbb C,n)$. Both spaces are vector spaces over $\mathbb R$ which have a natural topology induced by any norm. Thus $\text{Mat}(\mathbb R,n)$ is a both a linear and a topological subsapce of $\text{Mat}(\mathbb C,n)$, and $\text{Mat}(\mathbb R,n)$ is obviously closed in $\text{Mat}(\mathbb C,n)$.
But now $$O(n) = U(n) \cap \text{Mat}(\mathbb R,n)$$ which is a closed subset of $U(n)$.