This has little to do with the projective space.
Let $X$ be an algebraic variety over $\mathbb C$. It can be endowed with the natural topology induced by the complex absolute value. This topology is usually called the complex topology.
Lemma Let $Z$ be a complex algebraic variety. Let $T$ be a Zariski closed subset of $Z$, Zariski nowhere dense in $Z$. Then $T$ is complex nowhere dense in $Z$.
Let $Z^0$ be an open subset of a closed subset of $X$ (all in the sense of Zariski topology). We want to show
Claim: the Zariski closure of $Z^0$ coincides with its complex closure.
In your question, $X$ is a projective space, and $Z^0$ is an affine subvariety of $X$.
Proof: Let $Z$ be the Zariski closure of $Z^0$ and let $Z^c$ be the complex closure of $Z^0$. As the Zariski topology is coarse than the complex one, we have $Z^c\subseteq Z$. As $Z\setminus Z^0$ is Zariski closed and Zariski nowhere dense in $Z$, by the previous lemma, $Z\setminus Z^0$ is complex nowhere dense in $Z$, hence $Z^0$ is complex dens in $Z$. This implies that $Z^c=Z$.
It remains to prove the lemma. It is well-known, but I don't have a reference, so let me give a proof here. Shrinking $Z$ if necessary, we can suppose $Z$ is affine and Zariski closed in some $\mathbb C^n$. Let $I, J$ be the respective definining (radical) ideals in $\mathbb C[z_1,\dots, z_n]$ of $Z$ and $T$. By hypothesis, there exists
a complex open subset $U$ of $\mathbb C^n$ such that
$U\cap Z=U\cap T\ne\emptyset$.
We can suppose wlog that $p:=(0,..,0)\in U\cap T$. In the local ring $\mathcal O_{(\mathbb C^n)^{an},p}$ of germs of holomorphic functions, by analytic Nullstellensatz, we have $I \mathcal O_{(\mathbb C^n)^{an},p}=J\mathcal O_{(\mathbb C^n)^{an},p}$ because $U\cap Z=U\cap T$. Passing to the formal completion, we obtain
$$I\mathbb C[[z_1,\dots, z_n]]=J\mathbb C[[z_1,\dots, z_n]].$$
By the faithfull flatness of the formal completion $\mathbb C[z_1,\dots, z_n]_{\mathfrak m} \to \mathbb C[[z_1,\dots, z_n]]$ (where $\mathfrak m$ is the maximal ideal corresponding to $p$), we have
$$I\mathbb C[z_1,\dots, z_n]_{\mathfrak m}=J\mathbb C[z_1,\dots, z_n]_{\mathfrak m}.$$
This means that $T$ and $Z$ coincide in a Zariski open neighborhood of $p$. Contradiction with the hypothesis $T$ nowhere dense in $Z$.
Yes it is true. You probably know that a non-empty Zariski-open set $U\subseteq X$ is Zariski-dense, and it is a standard fact that $U$ is also Euclidean-dense. (See this mathoverflow answer for a slick proof of that fact.) Assuming this, the proof of your claim is simple:
Wlog, $Y$ is locally closed because the closure of a finite union is the union of the closures. Then, by definition, $Y$ is Zariski-open in $\bar{Y}^Z$ and hence $Y$ is Euclidean-dense in $\bar{Y}^Z$, i.e. $\bar{Y}^E\cap\bar{Y}^Z=\bar{Y}^Z$. Since $\bar{Y}^E\subseteq\bar{Y}^Z$, this concludes the proof.
Best Answer
This is going to sound crazy, but:
The reason is almost purely set-theoretic, and here it is.
People call a topological space $X$ a noetherian space if it satisfies the ascending chain condition on open sets (equivalently, the DCC on closed sets). Notice $\mathbb{A}^n$ is noetherian, in fact any variety (irreducible or not) is noetherian --- this is basically because the coordinate ring is a finitely-generated algebra over a field, which is a noetherian ring, and closed sets in $\mathbb{A}^n$ correspond to ideals of these rings.
Here is the important exercise.
Since $\mathbb{A}^n$ is noetherian, this answers your question!