I try to show that $\mathbb{R}^n$ is dense in $\mathbb{RP^n}$.
I can show that $$\mathbb{RP}^n=\mathbb{R}^n\cup\mathbb{R}^{n-1}...\cup\mathbb{R}^1\cup\mathbb{R}^0.$$
But now I have a question. Isn't the union of (for instance) $\mathbb{R}^3\cup\mathbb{R}^2=\mathbb{R}^3$? Because the lower dimensional space has to be included in the higher dimensional? Or can I say that the $\mathbb{R}^2$ has tupel $(x_0,x_1)$ and $\mathbb{R}^3$ has tupel $(x_0,x_1,x_2)$ and so they are different and the union is $\{(x_0,x_1),(x_0,x_1,x_2)\}$?
Edit:
Because there is the Question how I show the first equation.
I take a subspace $U_i\subset\mathbb{RP}^n$ with a fixed value for the i-th coordinate. $$U_i=\{(x_0:...:x_{i-1}:x_i=1:x_{i+1}:...:x_n)\}$$ all values of $x_j$ for $j\neq i$ are free, so $U_i$ is equivalent to $\mathbb{R}^n$.
Now i substract $U_i$ from $\mathbb{RP}^n$.
$$\mathbb{RP}^n\smallsetminus U_i=\{(x_0:...:x_{i-1}:0:x_{i+1}:...:x_n)\}$$ wich is $\mathbb{RP}^{n-1}$.
It follows that:
$$\mathbb{RP}^n=U_i\cup\mathbb{RP}^{n-1}.$$ I can repeat this until $\mathbb{RP}^0$.
Because $U_i$ is equivalent to $\mathbb{R}^n$, I can write $$\mathbb{RP}^n=\mathbb{R}^n\cup\mathbb{R}^{n-1}...\cup\mathbb{R}^1\cup\mathbb{R}^0.$$
Writing $\mathbb{RP}^n=\mathbb{R}^n\cup\mathbb{R}^{n-1}\cup...\cup\mathbb{R}^1\cup\mathbb{R}^0$ does not make sense because the $\mathbb R^i$ are certainly no subsets of $\mathbb{RP}^n$. But you can write $$\mathbb{RP}^n = U_n \cup U_{n-1}\cup...\cup U_1\cup U_0$$ where the $U_i$ are subspaces of $\mathbb{RP}^n$ which are homeomorphic to $\mathbb R^i$. These $U_i$ are pairwise disjoint, thus you never have $U_i \subset U_{i+1}$.
By the way, you should define $$U_i = \{(x_0: \ldots : x_{i-1}:x_i = 1:x_{i+1}=0: \ldots : x_n=0)\} .$$