How can I prove that the DE $y'=y^\alpha$ has infinitely many solutions?

Here how you can proceed. First note that $y=0$ is a solution. Next, you use integration to get $$ \int_0^y\frac{dy}{y^{\alpha}}=t+C_1. $$ Note that on the left hand side you have an improper integral which converges if $0<\alpha<1$ -- this is the reason the solutions are not unique at the point $(0,0)$. Assuming that $0<\alpha<1$ you get $$ y=K(t-C)^{\frac{1}{1-\alpha}} $$ also a solution to your equation. Here $K$ is a constant which depends on $\alpha$.

Here is a first example of a solution $$ y_1(t)=\begin{cases} 0&t<0,\\ Kt^{\frac{1}{1-\alpha}}&t\geq 0. \end{cases} $$ The only thing you need to check that it has a continuous derivative at $t=0$, which can be checked by direct calculation.

Generalizing, any function of the form $$ y_1(t)=\begin{cases} 0&t<C,\\ K(t-C)^{\frac{1}{1-\alpha}}&t\geq C. \end{cases} $$ is a solution (you need to check the derivative at $t=C$) and they all pass through $(0,0)$ if $C>0$.

You have $$ y' = \frac{dy}{dx} = y^{\alpha}, \ y(0)=0, \ 0<\alpha<1$$ From this we can start \begin{align} \frac{dy}{dx} &= y^{\alpha} \\ \Leftrightarrow \frac{1}{ y^{\alpha}} dy &= dx \\ \Leftrightarrow \int \frac{1}{ y^{\alpha}} dy &= \int 1 dx \end{align} Since this really does smell like homework like Siminore said, you should try from this point by your own. Remember the integration constant and fit that with the initial condition $y(0)=0$.

Edit: Okay if this isn't HW I shall continue.

Taking the above we get: \begin{align} \int \frac{1}{ y^{\alpha}} dy &= \int 1 dx \\ \Leftrightarrow \frac{y^{1-\alpha}}{1-\alpha}+c &=x \, \text{ as } \, 0<\alpha<1 \\ \Leftrightarrow y^{1-\alpha}&=(x-c)\cdot(1-\alpha) \\ \Leftrightarrow y &= ((x-c)\cdot(1-\alpha))^{\frac{1}{1-\alpha}} \end{align}

Now we also have to check whether our initial condition is fulfilled. $$ y(0) =((0-c)\cdot(1-\alpha))^{\frac{1}{1-\alpha}}=0$$ only for $c=0$, so $$ y(x) = (x\cdot(1-\alpha))^{\frac{1}{1-\alpha}} $$ is the unique solution of the above ODE.

Edit: It is not the unique solution, since $y(t)=0 \ \forall t$ also solves the problem.