Suppose that $\nabla^2 f$ is continuous in an open neighborhood of $x^*$ and that $\nabla f(x^*)=0$ and $\nabla^2 f(x^*)$ is positive definite. Then $x^*$ is a strict local minimizer of $f$
What I did:
Since $\nabla^2 f(x^*)>0$ and $\nabla^2 f$ is continuous in an open neighborhood of $x^*$, there exists $\overline{t}\in (0,T)$ such that $p^t\nabla^2 f(x^*+\overline{t}p)p>0$. Now, by the Taylor expansion of $f$ around $x^*+\overline{t}p$:
$$f(x^*+\overline{t}p) = f(x^*) + \nabla f(x^*) + \frac{1}{2}\overline{t}p^t\nabla^2f(x^*+\overline{t}tp)p$$
for some $t\in (0,1)$ which is the same as
$$f(x^*+\overline{t}p) = f(x^*) + \nabla f(x^*) + \frac{1}{2}\overline{t}p^t\nabla^2f(x^*+t'p)p$$
for some $t'\in (0,T)$, but we know that for any such $t'$, we have $\overline{t}p^t\nabla^2f(x^*+t'p)p>0$ and since $\nabla f(x^*)=0$,
$$f(x^*+\overline{t}p) - f(x^*) >0\implies f(x^*+\overline{t}p) > f(x^*)$$
This basically says that we can always find $z=x^* + \overline{t}p$ where $f(z)>f(x^*)$. It can be any $z$ in a ball around $x^*$ so I guess the proof is correct.
Can you spot any errors?
We want to prove that for every point $x$ in a small neighborhood around $x^*$ that is not equal to $x^*$, $f(x) > f(x^*)$ but your quantifier states that just there exists $\bar{t}\in (0,T)$ which need not be a point near $x^*$.
Since $\nabla^2 f(x^*)>0$ and $\nabla^2 f$ is continuous in an open neighborhood of $x^*$, $\forall p \in \mathbb{R}^n$, there exists $T>0$, such that for all $\overline{t}\in (0,T)$, we have $p^t\nabla^2 f(x^*+\overline{t}p)p>0$.