Embedding $RP^2$ into $R^4$

$f(\mathbf{x})=f(\mathbf{y}) \iff \mathbf{x} \sim \mathbf{y}$, where $\sim$ is the relation induced by $\mathbf{x} \sim \lambda \mathbf{x}$ for all $\lambda \in \mathbb R-\{0\}$ (or they are both zero)

Hence, you can show that it is injective, by the universal property of the quotient topology, which essentially says that it is a "final topology," the coarsest one that makes such a quotient continuous.

Hence, $f$ induces a unique continuous map $\tilde{f}:\mathbb R P^2 \to \mathbb R^4$, and it is injective by the first line.

I'm not sure that this avoids the use of "brute force," but it is easier to check that only elements "on the same line" will evaluate to the same vector under $f$.

See Section 2 Example of (Embedding of $RP^2$ into $R^4$): https://www.cs.uic.edu/~sxie/lecture_notes/diff_manifolds/lecture10.pdf