Given a field $k$, we define the scheme-theoretic $n$-th affine space over $k$ by $\mathbb{A}^n_k=\text{Spec}(k[X_1,\dots,X_n])$ and the $n$-th projective space over $k$ by $\mathbb{P}^n_k=\text{Proj}(k[X_0,\dots, X_n])$. We know $\mathbb{P}^n_k$ is covered by $n+1$ affine charts given by $D_+(X_i)=\mathbb{P}^n_k\smallsetminus V_+(X_i)$ for $i=0,\dots, n$, each isomorphic to $\mathbb{A}^n_k$.
I was asking myself if - as it happens in the naive case - each of those charts is dense in $\mathbb{P}^n_k$.
Here is my attempt. Take e.g. $D_+(X_0)$. Then $D_+(X_0)$ being dense in $\mathbb{P}^n_k$ is equivalent to $V_+(X_0)$ having empty interior. Suppose there exists $f\in (X_0,\dots, X_n)$ s.t. $D_+(f)\subset V_+(X_0)$. Then every prime $\mathfrak{p}\in \mathbb{P}^n_k$ s.t. $f\notin \mathfrak{p}$ is s.t. $X_0\in \mathfrak{p}$, which means $D_+(fX_0)=\emptyset$, i.e. $V_+(fX_0)=V_+(0)=\mathbb{P}^n_k$, and thus $f=0$ since $k[X_0,\dots, X_n]$ is an integral domain.
Is this proof correct? Does anyone know a shorter way to prove it?
Your argument is correct (you might just want to emphasize that the $f$ you introduce is homogeneous). The point is that the equality $V_+(fX_0)=\mathbf{P}_k^n$ is equivalent to $(fX_0)\cap(X_0,\ldots,X_n)\subseteq\sqrt{0}$; the lefthand side of this inclusion is just the ideal $(fX_0)$ since $fX_0\in(X_0,\ldots,X_n)$, while the righthand side is $0$ (the zero ideal) since $k[X_0,\ldots,X_n]$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $\mathbf{P}_k^n$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $D_+(X_i)$ and the manner in which they are glued together (more precisely, that $D_+(X_i)\cap D_+(X_j)\neq\emptyset$ for all $i,j$). In general, if you have a nonempty scheme $X$ that can be written as a union $\bigcup_{i\in I}X_i$ such that
(1) $I$ is nonempty,
(2) each $X_i$ is an irreducible open subset of $X$, and
(3) $X_i\cap X_j\neq\emptyset$ for all $i,j\in I$,
then $X$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $\mathrm{Proj}(S)$ for a wider class of graded rings $S$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.