I'm going to give you some examples. You can decide if they answer your proof or not.

Take $f(x, y) = x^2 + y^2$. Then $\nabla f(0, 0) = (0, 0)$, and
$$\nabla^2 f(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \succeq 0.$$
The fact that the matrix is positive definite means that it's also positive semi-definite. Note that $(0, 0)$ is a local (and in fact, global) minimum.

Take $g(x, y) = x^2$. Then $\nabla g(0, 0) = (0, 0)$, and
$$\nabla^2 g(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$
Note that $0$ is the minimal value of $g$, making $(0, 0)$ a local (and global) minimum. However, this minimum is achieved all along the line $x = 0$, e.g. at $(0, 1)$ as well.

Take $h(x, y) = x^2 + y^4$. Then $\nabla h (0, 0) = (0, 0)$, and
$$\nabla^2 h(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$
Again, we have a global minimum at $(x, y) = (0, 0)$, and this time it's unique. However, the Hessian was only positive semi-definite.

Finally, take $k(x, y) = x^2 - y^4$. Then $\nabla k(0, 0) = (0, 0)$, and
$$\nabla^2 k(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$
This time, we have positive semi-definiteness in the Hessian, but no local minimum.