The answer is in the negative and a construction of a counterexample is provided here. I wrote this argument quickly, but I hope it is correct. Let me know if it is not. We will construct a family of lines passing through the origin in $\mathbb{R}^3$ so that:
1. Every plane passing through the origin contains at least one of the lines from the family,
2. The union of the lines does not contain any plane passing through the origin.
The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the one used in this post.
Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that
(a) $\ell_\alpha\subset P_\alpha$,
(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$.
Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows.
The construction of the family of lines $\ell_\alpha$ is pretty standard. Suppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Consider the plane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines $P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).