Let $k$ be an algebraically closed field and $f: \mathbb{A}^2_k \rightarrow \mathbb{A}^2_k$, $(x,y) \mapsto (x,xy)$. I understand what the image looks like, i.e. $\mathbb{A^2_k} - (\mathbb{A}^1-0)$. By observing that: $\mathbb{A}^2_k = \textrm{Im} \cup \mathbb{A}^1_k$ and since $\mathbb{A}^2_k$ is irreducible, by taking the closure we get that $\overline{\textrm{Im}}$ is the whole plane and therefore that the image is not closed. If it was open, the intersection $\textrm{Im} \cap \{X \neq 0\} = 0$ would also be open, which is impossible because $\mathbb{A}^2_k$ is irreducible. Finally, since the image is dense and is not open in the whole set, it can't be locally closed either.
I find this reasoning to be a bit arbitrary. In particular, for the "locally closed" part, I initially wanted to look that at every neighbourhood of $\{(0,0)\}$, which would be the whole plane without some curves, but I feel like that is too tedious.
Is there a more intuitive (understand obvious) way to think about these things ? In general, is there an intuitive way to think about sets being closed or open in the Zariski topology?
Here is a different method than suggested in the comments. If the image is closed (resp. open), it's intersection with any set must also be closed (resp. open). Now pick some good sets to check against: the two axes will be instructive.