$\mathbb{CP}^1$ is compact? $\mathbb{CP}^1$ is the set of all one dimensional subspaces of $\mathbb{C}^2$, if $(z,w)\in \mathbb{C}^2$ be non zero , then its span is a point in $\mathbb{CP}^1$.let $U_0=\{[z:w]:z\neq 0\}$ and $U_1=\{[z:w]:w\neq 0\}$, $(z,w)\in \mathbb{C}^2$,and $[z:w]=[\lambda z:\lambda w],\lambda\in\mathbb{C}^{*}$ is a point in $\mathbb{CP}^1$, the map is $\phi_0:U_0\rightarrow\mathbb{C}$ defined  by $$\phi_0([z:w])=w/z$$ 
 the map  $\phi:U_1\rightarrow\mathbb{C}$ defined  by $$\phi_1([z:w])=z/w$$ 
Now,Could any one tell me why $\mathbb{CP}^1$ is the union of two closed sets $\phi_0^{-1}(D)$ and $\phi_1^{-1}(D)$, where $D$ is closed unit disk in $\mathbb{C}$,and why $\mathbb{CP}^1$ is compact?

The complex projective line is the union of the two open subsets
$$
D_i=\{[z_0:z_1]\text{ such that }z_i\neq0\},\quad i=0, 1.
$$
Each of them is homeomorphic to $\Bbb C$, hence to the open unit disk.
Compacteness follows from the fact that it is homeomorphic to the sphere $S^2$. You can see this as a simple elaboration of the observation that
$$
\Bbb P^1(\Bbb C)=D_0\cup[0:1].
$$
