How to perturb nonzero elements to get nonzero elements

Let $x \in \mathbb{R}^n$ be such that $x_i\neq 0$ for all $i\in \{1,\dots,n\}$. How can we perturb $x$ so that the perturbed vector $y$ has the same property, i.e., $y_i\neq 0$ for all $i\in \{1,\dots,n\}$?

My try:

Let $i^* \in \arg\min_{i \in \{1,\dots,n\}}|x_i|$. Then, there exists an open ball $B(x, \epsilon)$ where $\epsilon = \frac{|x_{i^*}|}{\gamma}$($\gamma>1$) and for all $y \in B(x, \epsilon)$ one has $y_i\neq 0$ for all $i\in \{1,\dots,n\}$. I know this is true intuitively, but I cannot show it. I need to show $0<|y_i-x_i|$ for all $i\in \{1,\dots,n\}$. We know $y=x+\epsilon v$ where $v \in \mathbb{R}^n$ with $\|v\|<1$ so $y_i=x_i+\epsilon v_i$. Can you please help me?

Using that $\,|\epsilon v_i| \le |\epsilon| \,\|v\| \lt |\epsilon| = \frac{|x_{i^*}|}{\gamma} \le \frac{|x_{i}|}{\gamma} \lt |x_i|\,$ and the triangle inequality:

$$ |y_i| \ge \big| |x_i|-|y_i-x_i| \big| = \big||x_i| - |\epsilon v_i| \big| = |x_i| - |\epsilon v_i| \gt |x_i| - |\epsilon| \gt 0 $$

Note: the question assumes that both inequalities $\,\|v\| \lt 1\,$ and $\,\gamma \gt 1\,$ are strict, though it is enough for one of them to be strict in order for the conclusion to hold.