$\mathbb{CP}^1$ 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 maps you gave are the coordinate charts on $\mathbb{C}\mathbb{P}^1$ that makes it into a manifold. In particular, they are bijective. 
If we take $\phi_0^{-1}(D)$, we get all point $[1:z]$, with $z\in D$. Similarly, $\phi_0^{-1}(D)$ is all points of the form $[z:1]$. Together, these sets cover $\mathbb{C}\mathbb{P}^1$. The inverse maps are continuous, because the maps $\phi_i$ are bijective, so $\phi_0^{-1}(D)$ and $\phi_1^{-1}(D)$ are compact, as the image of a compact set under a continuous map is compact. It's not hard to show that if a space is a union of two compact sets, it is compact, so we are done.
