Edit: According to the comment of L. Spice I changed the inclusion sign to the subset sign.
Is there a continuous map $f:\mathbb{C}P^3 \to \textrm{Gr}_{\mathbb{C}}(2,4)$ with $x\subset f(x)$? What about a map $g$ in the opposite direction with $g(x)\subset x$? What about a holomorphic version ($f$ or $g$ holomorphic)? What about a generalization about such maps between arbitrary grassmannian spaces?
I think there is no holomorphic such map. Consider the incidence variety $Z=\{(p,\ell)\in \mathbb{P}^3\times \mathbb{G}(2,4)\,|\, x\in\ell\} $. The projection $p:Z\rightarrow \mathbb{P}^{3}$ is a $\mathbb{P}^2$-bundle, in fact it is the projective tangent bundle to $\mathbb{P}^3$. You are asking for a section of this bundle; that would give a line bundle $M$ on $\mathbb{P}^3$ which is a subbundle of the tangent bundle $T_{\mathbb{P}^3}$. Computing $c_3$ one sees that this line bundle must be $\mathcal{O}_{\mathbb{P}^3}(2)$; but $H^0(T_{\mathbb{P}^3}(-2))$ is zero, so $\mathcal{O}_{\mathbb{P}^3}(2)$ does not inject into $T_{\mathbb{P}^3}$.
I do not know if there exists a continuous section (contrary to what I wrote before editing).
The map $f \mathrel: \ell \mapsto \ell \oplus \ell'$, where $\ell' = \mathbb C\cdot\overline{(b, -a, d, -c)}$ when $\ell = \mathbb C\cdot(a, b, c, d)$, satisfies your first condition.
(I originally had a version without the complex conjugation, which doesn't work because $\ell' = \ell$ when $\ell = \mathbb C\cdot(1, i, 1, i)$. Fortunately @AliTaghavi pointed out how to fix it. The candidate without the conjugation would have been holomorphic, hence contradicted @abx's answer.)
