How to choose $y(0)$ if $y'(x)=-\alpha y^2(x)/x^2$? Say we have a differential equation $\frac{dy}{dx}=-\alpha \frac{y^2}{x^2}$
This reduces to: $\int y^{-2}dy=-\alpha \int x^{-2} dx=-y^{-1}=-\alpha x^{-1} + C$. This can be rewritten as:
$$y= \frac{1}{C+\frac{\alpha}{x}}, \text{for } x\neq 0$$
$$\text {undefined, for } x=0$$
The second part is because according to the solution, neither $x$, nor $y$ can be equal to zero, since this would mean division by zero. 
It seems we have to leave $y$ undefined for $x = 0$ because the intermediate solution contained a division by zero of both $y$ and $x$. Nevertheless, $Lim_{x\to0}(y)=0$, so this clearly suggests that we do not have to leave $y$ undefined at $x=0$, but can simply set $y(0)=0$. 
For example here is a graph of the function with $C=\alpha=1$:

My question is: How do we rigorously justify setting $y(0)=0$, even though in the solution we are dividing by both $y$ and $x$, so that it should be undefined at $y=0,x=0$?

Your vector field is only well-defined for $x\neq 0$. The usual Cauchy Lipshitz condition is  not verified for $x=0$, so no guarantee a priori of existence/uniqueness of a solution through $x=0$.
It just happens so in this case, that the solutions from $x>0$ and $x<0$ may be joined as a $C^1$ function at $0$ when choosing the same const of integration on the two sides. For example if your ODE read $y'=-a y^2/x$, then the procedure wouldn't give rise to a $C^1$ function through 0, although it is continuous.
(there is btw a sign error in your solution).
