If we have a set $\mathcal L$ of lines in $\mathbb R^n$ such that $|\mathcal L|=\mathfrak c$, we can get the set $M$ with the desired properties using transfinite induction.

 * Take any well-ordering of the set of lines $\mathcal L=\{l_\alpha; \alpha<\mathfrak c\}$ (such that $\alpha\ne\beta$ implies $l_\alpha\ne l_\beta$.)
 * By transfinite induction we define $M_\alpha$ for $\alpha<\mathfrak c$ which is either singleton or empty set. If $l_\alpha$ intersects $\bigcup\limits_{\beta<\alpha} M_\beta$, we put $M_\alpha=\emptyset$. (We add nothing if $l_\alpha$ already contains some points.) Otherwise we put $M_\alpha=\{m_\alpha\}$ where $m_\alpha \in l_\alpha \setminus \bigcup\limits_{\beta<\alpha} l_\beta$. (This set is non-empty, since $l_\alpha$ has cardinality $\mathfrak c$ and the intersection $l_\alpha \cap \bigcup\limits_{\beta<\alpha} l_\beta$ has cardinality smaller than $\mathfrak c$; each of the lines $l_\beta$ can intersect $l_\alpha$ at most in one point.)
 * Now simply put $M=\bigcup\limits_{\alpha<\mathfrak c} M_\alpha$.

The set $M$ has the desired properties. Clearly, $M$ intersects each $l_\alpha$. Moreover, if $m\in M$ and $m=m_\alpha$, then $m$ is the only point on the line $l_\alpha$. (If $m$ was added in the $\alpha$'s step, then $l_\alpha$ does not contain any of the points from $\bigcup\limits_{\beta<\alpha} M_\beta$, i.e., the points added in the *preceding* steps. And the construction is done in such way that no new point on $l_\alpha$ can be added in the steps *after* $\alpha$.) 

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If $|\mathcal L|<\mathfrak c$ then we can simply choose one point from each set $l \setminus \bigcup\limits_{k\in \mathcal L} k$. This set is non-empty - it has cardinality $\mathfrak c$. 

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I should point out that Gerhard Paseman suggested [in a comment](https://mathoverflow.net/questions/278143/meeting-a-set-of-lines-in-mathbbrn#comment685389_278143) another solution using transfinite induction. And for $n=2$ he suggested in [another comment](https://mathoverflow.net/questions/278143/meeting-a-set-of-lines-in-mathbbrn#comment685387_278143) a solution which does not use transfinite induction.

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Proof of the existence of Mazurkiewicz two-point set is in somewhat similar vein, see also this question: [Subset of the plane that intersects every line exactly twice](https://mathoverflow.net/q/21470).