Let $\alpha, \beta$ be arbitrary elements of $\mathbb{R}$. Which of the following sets are intervals of $\mathbb{R}$? \begin{gather*} \{\mathrm{Re} \, z : z \in \mathbb{C}, |z - \alpha - i \beta| < 1 \}, \\ \{\mathrm{Im} \, z : z \in \mathbb{C}, |z - \alpha - i \beta| > 1 \}, \\ \{|z| : z \in \mathbb{C}, |z - \alpha - i \beta| \leqslant 1 \}, \\ \{|z|^2 : z \in \mathbb{C}, |z - \alpha - i \beta| \geqslant 1 \}. \end{gather*}
I think they all are intervals $\forall \, \alpha, \beta \in \mathbb{R}$, because these sets can be considered the image of a continuous map on a connected set of $\mathbb{C}$.
I am wondering if I can prove that without using topology facts.
The set $\{z \in \mathbb{C} : |z - \alpha - i \beta| = 1\}$ can be identified with the circle whith center $(\alpha, \beta)$ and radius $1$ in the plane $\mathbb{R}^2$. Drawing on this geometric intuition and using a little algebra, I can show that
\begin{gather*}
\{\mathrm{Re} \, z : z \in \mathbb{C}, |z - \alpha - i \beta| < 1 \} =\ ] \alpha - 1, \alpha + 1 [ \quad \text{and}\\
\{\mathrm{Im} \, z : z \in \mathbb{C}, |z - \alpha - i \beta| > 1 \} = \mathbb{R}.
\end{gather*}
What about the last two sets?
I will use polar coordinates for the disc's centre: $\alpha+\beta i=r_0e^{i\theta_0}$.
The third set means "take the absolute value of every point in the disc and map it onto the real line". If the disc doesn't contain the origin, it lies entirely within the circles $r=r_0\pm1$ and the set is bounded by the interval $[r_0-1,r_0+1]$. There are no gaps in the set since any number $s$ in the interval has a preimage $se^{i\theta_0}$ in the disc. Thus the set is an interval on $\mathbb{R}$. (If the disc does contain the origin, the argument is the same but with the interval being $[0,r_0+1]$.)
The fourth set uses the norm, a squared absolute value, and is an interval along the same lines. If the disc doesn't contain the origin, I may draw a line from there to infinity that is entirely in the set ($\theta=\theta_0+\pi$ suffices), yielding the interval $[0,\infty)$ (norm cannot be negative). If the origin is in the disc, $r=r_0-1$ is internally tangent to that disc, and drawing the same $\theta=-\theta_0$ line shows that the set is the interval $[(r_0-1)^2,\infty)$.