Suppose $E \subset \mathbb R^n$ is an open subset with $0 \notin E$. $E$ has following properties:(1) if $x = (x_1, \dots, x_n) \in E$, then $x(\alpha) = (\alpha x_1, \alpha^2 x_2, \dots, \alpha^n x_n) \in E$ for every positive real number $\alpha \in \mathbb R_+$ (2) if $y=(y_1, \dots, y_n) \in E$, then $y_i < 0$ for every $i =1, \dots, n$. I am wondering whether $E$ is connected.
My idea: Let $x, y \in E$. We first connect $x \to x(\alpha)$ and $y \to y(\beta)$ for $\alpha, \beta$ sufficiently small and then leverage the openness of $E$. But this does not work since even if $x(\alpha), y(\beta)$ are sufficiently close to $0$, we cannot guarantee they belong to the same norm ball.
Edit: I am trying to abstract an application problem. The second property was missing before. There was an answer below to show the set is not connected with only property $1$. Sorry to cause confusion.
For ease, I'm replacing your property (2) with $y_i > 0$ for each $i$ (which is the same property for all intents and purposes).
Take $n=2$ and let $E = E_1 \sqcup E_2$ for $E_1 = \{(\alpha x_1,\alpha^2 x_2) : x_1,x_2,\alpha > 0, \frac{x_2}{x_1^2} > 1\}$ and $E_2 = \{(\alpha x_1, \alpha^2 x_2) : x_1,x_2,\alpha > 0, \frac{x_2}{x_1^2} < 1\}$. The point is that the sets $E_1,E_2$ are well-defined: indeed, if $P := (\alpha x,\alpha^2 y)$ satisfies $\frac{y}{x^2} > 1$, then if we represented $P$ also as $(\beta \frac{\alpha x}{\beta}, \beta^2 \frac{\alpha^2 y}{\beta^2})$, then $\frac{\frac{\alpha^2 y}{\beta^2}}{(\frac{\alpha x}{\beta})^2} = \frac{y}{x^2}$. And clearly $E_1,E_2$ are disjoint open sets not containing $0$ that satisfy properties (1) and (2).