Can there be a non constant continuous function from the open unit disc $D=\{z\in\Bbb{C}~:~|z|<1\}$to $\Bbb{R}$ which takes only irrational values.
Here I couldn't apply intermediate value theorem as could be done if it were a real function to disprove the statement.
You can prove something a bit stronger, indeed this works if we replace the unit disc for any open path-connected set, or equivalently (in $\mathbb{C}$) a connected open set.
Let $f: U \subseteq \mathbb{C} \to \mathbb{R}$ be a continuous function with $U$ open and path connected. Let's fix $z,w \in U$ and a path $\gamma : [0,1] \to U$ from $z$ to $w$. Now, $f\gamma:[0,1] \to \mathbb{R}$ is continuous. Therefore, $f\gamma([0,1]) = f(\gamma[0,1]) \subseteq im(f)$ is connected, so $im(f)$ contains an interval and in particular, takes values in $\mathbb{Q}$.