Write $f^i:R^2-\{x_1,x_2\}\to R^2-\{x_i\}$ for the inclusion of the plane minus two points into the plane minus a single point. These induce maps $f_i*:{\pi }_1(R^2-\{x_1,x_2\})\to {\pi }_1(R^2-\{x_i\})$ on the fundamental groups of the twospaces. The original questionis asking for anon-zero element $\alpha$ in $\backslash Ker(f^1*,f^2*)$. However, since $R^2-\{x_1,x_2\}$ is homotopically equivalent to a one point union of two circles it has fundamental group isomorphic to the free group on two generators, say where $A$ corresponds to a loop around $x_1$ and $B$ to a loop around $x_2$. So, for example, Olof's picture above is ${\mathrm{AB}}^{-1}{A}^{-1}B$ and is therefore one of the four simplest possible examples (another is mine which is $A^{-1}{\mathrm{BAB}}^{-1}$). Now, writing ${\pi }_1(R^2-\{x_1\})=\langle A\rangle =Z$ and ${\pi }_1(R^2-\{x_2\}=\langle B\rangle =Z$ we get $f^1*(A^{n_1}{B}^{m_1}\dots A^{n_r}{B}^{m_r}=n_1+\dots +n_r$ and similarly $f^2*(A^{n_1}{B}^{m_1}\dots A^{n_r}{B}^{m_r})=m_1+\dots +m_r$. So the kernel of $(f^1*,f^2*)$ is just the elements where the sum of the powers of the $A$ components is 0 and the sum of the powers of the $B$ components is 0. This allows us to construct examples of arbitrary complexity. For example, $A^3{B}^2{A}^{-2}{B}^{-1}{A}^{-1}{B}^{-1}$ (see the picture below, I hope I drew it correctly). Well, I hope that was interesting.