I think this is obvious but I ask anyway.
Is $\mathbb{P}_r(\mathbb{C})$ quasi-compact in the Zariski topology? We already know the Zariski topology is not Hausdorff.
By the way, why is $\mathbb{P}_r(\mathbb{C})$ called the compactification of $\mathbb{C}^r$? Is this for the subspace topology?
Yes; $\mathbb P^r_{\mathbb C}$ is a finite union of quasi-compact spaces (namely $\mathbb C^r$'s).
Of course, this is not that interesting, and calling $\mathbb P^r_{\mathbb C}$ a compactification seems not to make sense.
In one sense, if using usual Euclidean topology, then $\mathbb P^r_{\mathbb C}$ is indeed compact whereas $\mathbb C^r$ is not. Easy example: $\mathbb P^1_{\mathbb C}$ is homeomorphic to the sphere, which is a one-point compactification of $\mathbb R^2 \simeq \mathbb C$ by adding on "the point at infinity".
Another feel for compactification: Among the key property of $\mathbb P^r$ that is not true for $\mathbb C^r$ is that subvarieties not too small do intersect; e.g. any two lines in $\mathbb P^2$ intersect, whereas they don't necessarily in $\mathbb C^2$. It is in this sense that $\mathbb P^r$ is "compactified" because we have added in all the "points at infinity."
So, (at least if working over $\mathbb C$), the word "compactification" makes sense in that way.
The real notion lurking in the background is that of "properness" (in scheme theoretic language), which essentially means in our case the following: A $\mathbb C$-variety $X$ is proper if $X\times_{\mathbb C} Y \to Y$ is a closed map for any other variety $Y$ (this is in fact just an analogue of what happens in Hausdorff case when $X$ is compact).