$ \mathbb{P}^n=\bigcup_{j=1}^{n} U_j,$ where $U_j=\{[x_0:\dots,x_n]\in \mathbb{P}^n : x_j\neq 0\}=\{[x_0:\dots,x_n]\in \mathbb{P}^n : x_j=1\}$ can be identified with $\mathbb{C}^n.$ In fact we need not use the complements of the hyperplanes defined by $x_i$ to define a cover $\{U_i\}$.Given any set of $n+1$ linearly independent hyperplanes not passing through the origin, the lines in $\mathbb{C}^{n+1}$ intersecting the i'th hyperplane form a set $U_i$ that can be identified with $\mathbb{C}^n,$ and together these give a cover of $\mathbb{P}^n$ that differs from the one we first described only by a change of coordinates in $\mathbb{C}^{n+1}.$ Note that there is a natural Eucidean topology on $\mathbb{P}^n$ induced by virtue of the fact that $\mathbb{P}^n$ is a quotient of $\mathbb{C}^{n+1}\setminus\{0\}$. In particular, two points of $\mathbb{P}^n$ are close together if the corresponding lines in $\mathbb{C}^{n+1}$ have small angle between them. in this Eucledian topology on $\mathbb{P}^n$, each of the sets $U_i$ is open, and identification of $U_i$ with $\mathbb{C}^n$ descriebd above defines a homeomorphism of topological spaces when $\mathbb{C}^n$ considered with its Euclidean topology.
Why each $U_i$ is dense in $\mathbb{P}^n?$ and in fact, its complement is a lower dimensional space (namely, $\mathbb{P}^{n-1}$.)The intersection of $U_i$ with $U_j$ when $i\neq j$ is also dense.