Consider a unitary group $U(n)=\{A\in GL_n(\mathbb{C}):A^*A=I\}$. I want to show that $U(n)$ is a compact group. First, I can observe that $U(n)\subseteq M_n(\mathbb{C})$ i.e. it's a subspace of an Euclidian space $M_n(\mathbb{C})$ with a metric given by $d(A,B)=||A-B||$ where $||A||=\sqrt{\text{tr}(A^*A)}$ where $M_n(\mathbb{C})$ is naturally isomorphic to $\mathbb{C}^{n^2}$ where $\mathbb{C}^{n^2}\cong \mathbb{R}^{n^2}\oplus \mathbb{R}^{n^2}$. So, it's enough to show that $U(n)$ is closed and bounded.
To show that $U(n)$ is bounded, it's enough to show that the norm of all elements of $U(n)$ is bounded by some constant. Indeed, observe $$||A||=\sqrt{\text{tr}(A^*A)}=\sqrt{\text{tr}(I)}=\sqrt{n}.$$ So, we can see that $||A||\leq n$ for all $A\in U(n)$. Therefore, $U(n)$ is bounded.
Next, let's show that $U(n)$ is closed. I want to take a direct approach. To show that $U(n)$ is closed, it's enough to show that it contains all its limit points. Indeed, consider some sequence of unitary matrices $\{A_n\}$ that converges to $A$. We want to show that $A\in U(n)$ i.e. $$A^*A=I.$$ But, it follows from the following sequence of equalities $$I=\lim_{n\to\infty}I=\lim_{n\to\infty}A^*_nA_n=\lim_{n\to\infty}(B^t_n-iC^t_n)(B_n+iC_n)=\lim_{n\to\infty}(B^t_nB_n+C^t_nC_n)+i\lim_{n\to\infty}(B^t_nC_n-C^t_nB_n)=(*).$$ Where using the fact that the limit of a matrix is the limit of each individual component and the product rule, we will get that $(*)=A^*A$. Can I use this direct approach? Did I miss some small details?