There is a solution using the axiom of choice, similar to this. Enumerate the points of space as $\{p_\alpha:\alpha<\phi\}$ where $\phi$ is the least ordinal of cardinality continuum. This means that we have a well ordering of $\mathbb{R}^3$ such that each element is preceded by less than continuum many elements. We are going to choose, by transfinite recursion on $\alpha$, a line $L_\alpha$ throu $p_\alpha$ such that these lines are not parallel and cover the space. Actually, for some $\alpha$ we do not choose $L_\alpha$.

Assume that we have made the choices for all $\beta<\alpha$ and we have to treat $p_\alpha$. We do nothing, if some earlier $L_\beta$ covers $p_\alpha$. Otherwise, draw a sphere $S$ around $p_\alpha$. We have to select a line $L_\alpha$ throu $p_\alpha$ and this is the same as to choose a point of $S$. Those points are disqualified which give rise to lines parallel to some earlier lines, this gives a set of less than continuum points on $S$. Further, to any given line $L_\beta$ ($\beta<\alpha$) we cannot choose any line that intersects $L_\beta$. The directions corresponding to these bad choices form a great circle on $S$. We have, therefore, a collection of less than continuum many points and great circles on $S$. An easy argument shows that they cannot cover $S$ [choose a great circle $C$ different from them, then the pints/great circles can cover at most two points of $C$ each], so we can make the choice of the diection of $L_\alpha$.

There is a solution using the axiom of choice, similar to this. Enumerate the points of space as $\{p_\alpha:\alpha<\phi\}$ where $\phi$ is the least ordinal of cardinality continuum. This means that we have a well ordering of $\mathbb{R}^3$ such that each element is preceded by less than continuum many elements. We are going to choose, by transfinite recursion on $\alpha$, a line $L_\alpha$ through $p_\alpha$ such that these lines are not parallel and cover the space. Actually, for some $\alpha$ we do not choose $L_\alpha$.

Assume that we have made the choices for all $\beta<\alpha$ and we have to treat $p_\alpha$. We do nothing, if some earlier $L_\beta$ covers $p_\alpha$. Otherwise, draw a sphere $S$ around $p_\alpha$. We have to select a line $L_\alpha$ throu $p_\alpha$ and this is the same as to choose a point of $S$. Those points are disqualified which give rise to lines parallel to some earlier lines, this gives a set of less than continuum points on $S$. Further, to any given line $L_\beta$ ($\beta<\alpha$) we cannot choose any line that intersects $L_\beta$. The directions corresponding to these bad choices form a great circle on $S$. We have, therefore, a collection of less than continuum many points and great circles on $S$. An easy argument shows that they cannot cover $S$ [choose a great circle $C$ different from them, then the pints/great circles can cover at most two points of $C$ each], so we can make the choice of the direction of $L_\alpha$.

There is a solution using the axiom of choice, similar to this. Enumerate the points of space as $\{p_\alpha:\alpha<\phi\}$ where $\phi$ is the least ordinal of cardinality continuum. This means that we have a well ordering of $\mathbb{R}^3$ such that each element is preceded by less than continuum many elements. We are going to choose, by transfinite recursion on $\alpha$, a line $L_\alpha$ throu $p_\alpha$ such that these lines are not parallel and cover the space. Actually, for some $\alpha$ we do not choose $L_\alpha$.

Assume that we have made the choices for all $\beta<\alpha$ and we have to treat $p_\alpha$. We do nothing, if some earlier $L_\beta$ covers $p_\alpha$. Otherwise, draw a sphere $S$ around $p_\alpha$. We have to select a line $L_\alpha$ throu $p_\alpha$ and this is the same as to choose a point of $S$. Those points are disqualified which give rise to lines parallel to some earlier lines, this gives a set of less than continuum points on $S$. Further, to any given line $L_\beta$ ($\beta<\alpha$) we cannot choose any line that intersects $L_\beta$. The directions corresponding to these bad choices form a great circle on $S$. We have, therefore, a collection of less than continuum many points and great circles on $S$. An easy argument shows that they cannot cover $S$ [choose a great circle $C$ different from them, then the pints/great circles can cover at most two points of $C$ each], so we can make the choice of the diection of $L_\alpha$.

There is a solution using the axiom of choice, similar to this. Enumerate the points of space as $\{p_\alpha:\alpha<\phi\}$ where $\phi$ is the least ordinal of cardinality continuum. This means that we have a well ordering of $\mathbb{R}^3$ such that each element is preceded by less than continuum many elements. We are going to choose, by transfinite recursion on $\alpha$, a line $L_\alpha$ throu $p_\alpha$ such that these lines are not parallel and cover the space. Actually, for some $\alpha$ we do not choose $L_\alpha$.

Assume that we have made the choices for all $\beta<\alpha$ and we have to treat $p_\alpha$. We do nothing, if some earlier $L_\beta$ covers $p_\alpha$. Otherwise, draw a sphere $S$ around $p_\alpha$. We have to select a line $L_\alpha$ throu $p_\alpha$ and this is the same as to choose a point of $S$. Those points are disqualified which give rise to lines parallel to some earlier lines, this gives a set of less than continuum points on $S$. Further, to any given line $L_\beta$ ($\beta<\alpha$) we cannot choose any line that intersects $L_\beta$. The directions corresponding to these bad choices form a great circle on $S$. We have, therefore, a collection of less than continuum many points and great circles on $S$. An easy argument shows that they cannot cover $S$ [choose a great circle $C$ different from them, then the pints/great circles can cover at most two points of $C$ each], so we can make the choice of the diection of $L_\alpha$.