Question: Let $X$ be an affine variety over $\Bbb C$, and let $Y\subseteq X$ be a constructible set (i.e. $Y$ is a finite union of locally closed sets). Is it true that the Zariski closure of $Y$ is the same as the closure of $Y$ in the standard Euclidean topology inherited from the inclusion $X\subseteq\Bbb C^n$?
In this question I asked whether these two closures were the same when $Y$ is the orbit of an algebraic group action. Then, this answer says that the answer is yes because orbits are constructible sets. However, I don't know a proof of the fact that these orbit closures are the same for constructible sets.
If $\bar{Y}^E$ denotes the euclidean closure and $\bar{Y}^Z$ the Zariski one, then it is clear that $$\bar{Y}^E\subseteq \bar{Y}^Z$$ since the Euclidean topology is finer than the Zariski topology. But for the converse, I am clueless.