If $|z| \leq 1$, then $|\text{Im}(\bar{z} + 1 )| \leq \sqrt{3-x^2}$ If $|z| \leq 1$, then $|\text{Im}(\bar{z} + 1 )| \leq \sqrt{3-x^2}$ $\quad $ for $0 \leq x \leq1.$
Attempt:
Let $z = x + iy$ such that $\bar{z} = x -iy$.
Thus,
$$|z|^2 = x^2 + y^2 \leq 1 \quad \quad \quad (1)$$ and
$$\bar{z} = x - iy$$
$$\bar{z} + 1 = (x+1) - iy$$
$$\text{Im}(\bar{z} + 1) = -y$$
$$|\text{Im}(\bar{z} + 1 )| = |-y| = |y| \leq \sqrt{1-x^2} \leq \sqrt{3-x^2}\quad \quad \text{(because of the inequality in (1))}$$
which is true for $0 \leq x \leq1.$
Are my arguments until the final answer valid and logical?

Your way is fine and correct, more simply, as an alternative, from here $|y| \leq \sqrt{3-x^2}$ we can square both side to obtain
$$y^2\le 3-x^2 \iff x^2+y^2 \le 3$$
which is true since  $x^2 + y^2 \leq 1$.

Another way, by polar coordinates with $|r|\le 1$
$$|y| \leq \sqrt{3-x^2} \iff r|\sin \theta|\le\sqrt{3-r^2\cos^2 \theta}\le\sqrt 2$$
which is true since $r|\sin \theta|\le 1$.
