Prove: if $f(0)=0$ and $f'(0)=0$ then $f''(0)\geq 0$
Your proof is enough when $f''$ is continuous.
Here's a way without the continuity assumption.
This Taylor expansion $f(x) = f(0) + xf'(0) + \frac{x^2}{2}f''(0) + o(x^2)$ yields here: $$f(x) = \frac{x^2}{2}f''(0) + o(x^2)$$
Either $f''(0)=0$ and we're good, otherwise the previous equality rewrites as $\displaystyle \lim_{x\to 0}\frac{2f(x)}{f''(0) x^2}= 1$
Consequently, $f(x)$ and $f''(0) x^2$ must share the same sign on a neighborhood of $0$.
Since $f\geq 0$ and $x^2\geq 0$ that implies $f''(0)\geq 0$
When $f''(0)=0$, $f(x)=o(x^2)$. I don't see anything more you could say.
Another approach is via the method of contradiction. Assume that $f''(0) < 0$. Then the function $f'$ is strictly decreasing at $0$ which means that there is a neighborhood $I$ of $0$ such that if $x \in I, x < 0$ then $f'(x) > f'(0) = 0$ and if $x \in I, x > 0$ then $f'(x) < f'(0) = 0$. We can obviously choose $I$ of the form $(-h, h)$ and hence observing sign of $f'$ in $(-h, h)$ we see that $f$ is strictly increasing in $(-h, 0]$ and strictly decreasing in $[0, h]$ and since $f(0) = 0$ it follows that $f(x) < 0$ for all $x \in I$ except $x = 0$. And this is contrary to the hypotheses that $f$ is non-negative. Thus we obtain the desired contradiction.
BTW the above argument can be replaced by the following concise argument. Since $f'(0) = 0, f''(0) < 0$ the point $0$ is a local strict maximum of $f$ and hence $f(x) < f(0) = 0$ for all $x$ in some neighborhood of $I$ of $0$ except $x = 0$. And this contradicts that $f$ is non-negative.
The argument in first paragraph actually shows how the vanishing of derivative and sign of second derivative guarantee local maxima/minima and if the reader is well aware of the proof of second derivative test for maxima/minima, it is preferable to adopt the argument in second paragraph.
As can be seen from the arguments above we don't need continuity of $f''$. And we can avoid difficult theorems like Taylor or L'Hospital. The argument used in first paragraph leads to one of the simpler proofs of Taylor's Theorem with Peano's form of remainder.
Update: The condition $f' (0)=0$ is superfluous here. The function is non-negative and $f(0)=0$ and hence $0$ is a point of local minima so that $f'(0)=0$.
For 2) you could have $f(x)=x^4$ which satisfies all properties and conclusions including $f''(0)=0$, yet $f(x)$ is not identically zero.
For 1) relaxing the condition that $f(0)=0$, we could look at $f(x)=\cos(x)+3$ (which has instead $f(0)=4$). It satisfies the property that $f(x)$ is non-negative, is twice differentiable on $[-1,1]$ and that $f'(0)=0$. However, $f''(0)=-\cos(0)=-1$ is not non-negative.
Relaxing the condition that $f(x)$ be non-negative on the interval, letting $f(x)=-x^2$ one has $f(0)=0, f'(0)=0$, yet $f''(0)=-2$ is not non-negative
Relaxing the condition that $f$ be twice differentiable makes no sense since then $f''(0)$ can't be talked about.
Try searching yourself as well for a function $f(x)$ which violates only the hypothesis that $f'(0)=0$ which has $f''(0)<0$.
With these counterexamples, we conclude that in fact every hypothesis was necessary.