Triviality of a tautological bundle

Consider a section $s$ of this line bundle. Let $\phi\colon [0,1] \to S^n$ be a path from a point $p\in S^n$ to $-p$.

We have a unique continuous function $\lambda\colon [0,1]\to \mathbb{R}$ such that $s([\phi(t)])=\lambda(t)\phi(t)$ for all $t\in[0,1]$.

We have: $$\lambda(0)p=s([\phi(0)])=s([\phi(1)])=-\lambda(1)p$$

Thus $\lambda(0)=-\lambda(1)$ and by the intermediate value theorem $\lambda(t)=0$ for some $t\in [0,1]$.

Thus this bundle has no non-vanishing sections, and is different to the trivial line bundle.

Let $f : \mathcal{T} \to \mathbb{R}P^n \times \mathbb{R}$ be a bundle isomorphism. The quotient map $\mathbb{R}^{n + 1} \to \mathbb{R}P^n$ is continuous, so we have a continuous inclusion $$ \iota : \mathbb{R}^{n + 1} \setminus \{0\} \to \mathbb{R}P^n \times \mathbb{R}^{n + 1} $$ which does the quotient map on the first factor and is the identity on the second factor. If we restrict the codomain of $\iota$ to the complement of the image of the zero section $\pi^{-1}(0)$ in $\mathcal{T}$, then $\iota$ is a homeomorphism $\mathbb{R}^{n + 1} \setminus \{0\} \to \mathcal{T} \setminus \pi^{-1}(0)$. Similarly $f$ restricts to a homeomorphism $$ f : \mathcal{T} \setminus \pi^{-1}(0) \to \mathbb{R}P^n \times (\mathbb{R} \setminus \{0\}). $$ Composing these two maps gives a homeomorphism $f \circ \iota : \mathbb{R}^{n + 1} \setminus \{0\} \to \mathbb{R}P^n \times (\mathbb{R} \setminus \{0\})$. But of course the former space is connected, while the latter is not: hence the map $f$ cannot exist.