Non ce la fa proprio!
Yes, that is correct. While it is true that a continuous map sends closed sets to closed sets in general, there are exceptions to this rule, and there are examples of continuous functions that do not have this property. Therefore, it is not true that a continuous map always sends closed sets to closed sets.
To see why this is the case, we can consider the function $f : \mathbb{R} \to [0,1)$ defined by $x \mapsto e^{-|x|}$. This function is continuous, which means that it satisfies the property that for any open set $U$ in the codomain space $[0,1)$, the inverse image of $U$ under the function $f$ is an open set in the domain space $\mathbb{R}$.
However, as we have seen above, the function $f$ does not have the property that it maps closed sets to closed sets. This is because the interval $[0,1)$ is not a closed set, and therefore the inverse image of the interval $[0,1)$ under the function $f$ is not a closed set in the domain space $\mathbb{R}$.
Therefore, while it is true that a continuous map sends closed sets to closed sets in general, there are exceptions to this rule, and there are examples of continuous functions that do not have this property. These exceptions are important because they tell us something about the topological properties of the function, and they can be used to study the behavior and properties of the function in more detail.