Proving ${\mathbb{P}}^n$ is Hausdorff

Your proof of $(1)$ is correct. In fact, you can generalize to this lemma:

A space $X$ is Hausdorff if for any two $x, x' \in X$ such that $x \ne x'$ there exists a continuous map $f : X \to Y$ to a Hausdorff space $Y$ such that $f(x) \ne f(x')$.

Note that $Y$ and $f$ are allowed to depend on $x,x'$. Also the converse is true (simply take $Y = X$ and $f = id$ for all $x,x'$).

Concerning $(2)$: The construction of $f$ does not make sense in the present form. Writing $f([\nu])$ means that you want to define it for the elements of $\mathbb P^n$. For each $\omega \in \mathbb R^{n+1} \setminus \{0\}$ define

$$ f_\omega :\mathbb R^{n+1} \setminus \{0\} \to \mathbb R, f_\omega(\nu) = \lvert \omega \rvert^2 \left(1- \frac{(\omega \cdot \nu)^2}{\lvert \omega \rvert^2 \lvert \nu \rvert^2}\right) .$$ It is obvious that if $\nu \sim \nu'$, then $f_\omega(\nu) = f_\omega(\nu')$. Thus $f$ induces a unique function $F_\omega : \mathbb P^n \to \mathbb R$ such that $f_\omega = F_\omega \circ \pi$. By the universal property of the quotient topology $F_\omega$ is continuous. Note that $F_\omega([\omega]) = 0$.

Now consider any two distinct points $[\nu],[\omega]$ in $\mathbb P^n$. This means that $\nu$ does not lie on the line $\mathbb R \omega$, thus $(\omega \cdot \nu)^2 < \lvert \omega \rvert^2 \lvert \nu \rvert^2$ (Cauchy-Schwarz). Therefore $F_\omega ([\nu]) \ne 0 = F_\omega([\omega])$. The function $F_\omega$ depends on $[\omega]$, but this is allowed.

Remark:

We could also work with $f_\omega(\nu) =\frac{(\omega \cdot \nu)^2}{\lvert \omega \rvert^2 \lvert \nu \rvert^2}$. This is the square of the cosine of the angle between $\omega$ and $\nu$. Then $F_\omega ([\omega]) = 1$ and $F_\omega ([\nu]) < 1$.