Does the metric $d(x,y)=||x-y||^p$ for $0Let $\| \cdot \|$ be the Euclidean norm on $\mathbb{R}^n$,  $0<p<1$ and $d: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ be the function defined by
$$d(x,y)=||x-y||^p, \quad x,y \in \mathbb{R}^n. $$
I have demostrated that $d$ is a metric on $\mathbb{R}^n$. Does $d(x,y)$ induce the usual topology on $\mathbb{R}^n$?

Yes, it does generate the same topology. The reason? The balls are the same! If $B_1(x, r)$ is the open ball centred at $x$, radius $r$, in this metric space, while $B_2(x, r)$ is the corresponding open ball with respect to the usual Euclidean metric, then
$$B_2(x, r) = B_1(x, r^p),$$
as
$$\|x - y\| < r \iff \|x - y\|^p < r^p \iff d(x, y) < r^p.$$
