I want to understand the topology/graph of the solution set $X$ of the equation $z_0^2+z_1^2+z_2^2=0$ in $\Bbb{C}P^2$.
If $z_0=0$ then $z_1^2+z_2^2=0$ gives only two points as solution: $P_{+}=[0,1,i]$ and $P_{-}=[0,1,-i].$ If $z_0\neq0$, we have $W_{+}=\{[1,z,\sqrt{-1-z^2}]\;|\; z\in\Bbb{C}\}$ and $W_{-}=\{[1,z,-\sqrt{-1-z^2}]\;|\; z\in\Bbb{C}\}$ as solution sets and $W_{+}\cap W_{-}=\{Q_1,Q_2\}$ where $Q_1=[1,i,0]$ and $Q_2=[1,-i,0]$. When $z\rightarrow\infty$, we observe that $W_{+}$ and $W_{-}$ are compactifed by $P_{+}$ and $P_{-}$ respectively.
Hence, I deduced that $X\simeq S^2\lor\lor S^2$ where $\lor\lor$ means that two spheres are joined at two different points say $Q_1$ and $Q_2$. Hence, $H_0(X)=\Bbb{Z}$, $H_1(X)=\Bbb{Z}$ and $H_2(X)=\Bbb{Z}\oplus\Bbb{Z}$. So, $\pi_1(X)=\Bbb{Z}$ and $X$ is not simply connected. The Euler characteristic of $X$ is $2$.
Where am I going wrong? Are there singularities here in my solution? Is $\pi_1(X)=\Bbb{Z}_2?$ I heard so.
I'm not certain of where you're going wrong, but a guess is that it's in using $\pm \sqrt{-1-z^2}$ - there may be some branch cut related issues.
Instead, here's another approach.
Proposition: The space $X$ is homeomorphic to the Grassmannian of oriented $2$-planes in $\mathbb{R}^3$. (More generally, repeating this in $\mathbb{C}P^n$ gives the oriented $2$-planes in $\mathbb{R}^{n+1}$.)
Proof: I will call the Grassmannian $G$ to save on typing.
Writing each $z_k= x_k + iy_k$ with $x_k,y_k$ real, then $X$ is the the zero set of $\sum_k(x_k^2 - y_k^2) + 2i\sum_k x_k y_k$. Taking real and imaginary parts, we deduce that $(z_1,z_2,z_3)\in X$ iff the real vectors $V_x:=\langle x_1,x_2,x_3\rangle$ and $V_y:=\langle y_1,y_2,y_3\rangle$ have the same length and are perpendicular.
Define a map $f:X\rightarrow G$ by $f(z_1,z_2,z_3) = \operatorname{span}\{V_x, V_y\}$ oriented so that $V_x,V_y$ is an oriented basis. Let's check that this is well defined. First, if we scale $(z_1,z_2,z_3)$ by a real number, this scales $V_x$ and $V_y$, but doesn't change their span. Second, if we multiply $(z_1,z_2,z_3)$ by $e^{i\theta}$, the net effect is that $V_x$ is replaced by $\cos(\theta)V_x + \sin(\theta) V_y$ and $V_y$ is replaced by $-\sin(\theta)V_x + \cos(\theta) V_y$. In particular, the oriented span doesn't change.
Is $f$ surjective? Well, given an oriented $2$-plane $P\subseteq \mathbb{R}^3$. Pick an oriented orthonormal basis $\langle x_1,x_2,x_3\rangle$ and $\langle y_1,y_2, y_3\rangle$ of $P$. Then setting $z_k = x_k + iy_k$, it's easy to verify that $(z_1,z_2,z_3)\in X$ and that $f(z_1,z_2,z_3) = P$.
Is $f$ injective? Well, if $f(z_1,z_2,z_3) = P = f(z_1', z_2', z_3')$, then we may obviously rescale both $(z_1,z_2,z_3)$ and $(z_1',z_2',z_3')$ to have length $2$ (so that the vectors $V_x$, etc are unit length).
Then $V_x$ and $V_y$ form an oriented orthonormal basis for $P$, as do $V_{x'}$ and $V_{y'}$. As any two such bases are related by a rotation by some angle $\theta$, it's now easy to verify that $z_k' = e^{i\theta} z_k$ for all $k$, so that $f$ is injective. $\square$
Alright, so we now know that $X$ is homeomorphic to $G$. But what is the topology of $G$? Well, in the special case you're interested in, $G$ has a fairly simple topology.
Proposition: The Grassmannian of oriented $2$-planes in $\mathbb{R}^3$ is homeomorphic to a sphere $S^2$.
Proof: Given a $2$-plane $P\in G$, there are precisely two unit vectors which are perpendicular to $P$. If we consider the orientation induced on $\mathbb{R}^3$ by taking the orientation on $P$ and following it with one of these two unit vectors, precisely one choice, denoted $V_p$ will return the orientation on $\mathbb{R}^3$. We define $f:G\rightarrow S^2$ by $f(P) = v_p$.
I will leave it to you to verify that this is a homeomoprhism. $\square$