I have a case where I have a linear regression model: $$y = X\beta^* + w$$ where $y \in \mathbb{R}^{n}$, $X \in \mathbb{R}^{n \times 1}$, $\beta^* \in \mathbb{R}^{1}$ and $w \sim N(0,I_n), w \in \mathbb{R}^{n}$ and $|\beta^*| \ge 1$ I want to prove that: $$\|\hat{\beta} - \beta^*\|_2^2 = O(\|\hat{\beta}\|^2_2)$$
Where $\hat{\beta}$ is least squares estimator of $\beta^*$ (formula is $\hat{\beta} = (X^TX)^{-1}X^Ty$) and that $|\hat{\beta}| \ge \frac{1}{2}$ is true with probability $1 - O(d^{-10})$ of time.
Can I boldly assume that $\|\hat{\beta} - \beta^*\|_2^2 = O(\|\hat{\beta}\|^2_2)$ is right from looking at the $\beta^*$ as a constant and by extending the square binomial: $$\|\hat{\beta} - \beta^*\|_2^2 = \|\hat{\beta}\|^2_2 - 2\langle \hat{\beta}, \beta^*\rangle + \|\beta^*\|^2_2 = O(\|\hat{\beta}\|^2_2)$$
Assuming that $X$ is non-random, $\hat{\beta}$ is a sequence converging to $\beta^*$ a.s. In that case, your argumentation looks fine. A small step that you omit is that $$| <\hat{\beta}, \beta^*>|\leq ||\hat{\beta}||\cdot ||\beta^*|| = O(||\hat{\beta}||^2),$$ by Cauchy-Schwarz, with the $O(||\hat{\beta}||^2)$ following from the fact that $\beta^*\neq 0.$