Prove that three skew (i.e. non-intersecting) lines in $P^{3}$(R) have an infinite number of transversals (i.e. lines meeting all three). So I think I might need to use the concept of general position but not sure how to proceed. What is a good way to prove this? Thanks.
 A: Let $L_i$ be the lines.
Take $P_1\in L_1$ and consider the unique plane $\Pi$ passing through $P_1$ and containing $L_2$.
 That plane does not  contain $L_3$ : else it would contain both $L_2$ and $L_3$  and since two lines in a projective plane always intersect, $L_2$ would intersect $L_3$, contrary to the skewness assertion.
So $\Pi$ cuts $L_3$ in just one point $P_3$ and the line $\overline {P_1P_3}$ which evidently cuts $L_1$ and $L_3$ also cuts $L_2$, since $\overline {P_1P_3}, L_2$ are both lines in $\Pi$.    
Conclusion
The line  $\overline {P_1P_3}$ cuts all $L_i$,  as required.
 By varying $P_1$ we obtain infinitely many lines $M_{P_1}$ cutting the original three $L_i$'s.   
Edit: a (very optional)  complementary remark
The lines $M_P$ described above have as union a quadric $\mathcal Q=\cup_{P\in L_1}M_P\subset \mathbb P^3(\mathbb R)$.
The set of disjoint lines $M_P$ partition $\mathcal Q$ into what is called a ruling of $\mathcal Q$ and the three lines $L_i$ are part of another ruling of $Q$.
Through each point $A\in \mathcal Q$ there passes one line from each ruling: $L_A, M_A$ and their union is the intersection of the quadric with its tangent plane at $A$: $$  L_A \cup  M_A  =\mathbb T_A(\mathcal Q) \cap \mathcal Q $$
Here is a concrete example [where the $M_{(p:q)}$ with $(p:q)\in \mathbb P^1(\mathbb R)$ form one ruling], left as an exercise or as a token of trusssst in me: $$\mathcal Q=\{xw-yz=0\}, L_1=\{z=w=0\}, L_2=\{x=z, y=w\}, \\
 L_3=\{x=y=0\} \; M_{(p:q)}=\{qx=py,qz=pw\}$$
