We use the following result, which is a consequence of Fubini's theorem:
$$\mu(\{x\})=\lim_{T\to +\infty}\frac 1{2T}\int_{-T}^Te^{-itx}\widehat\mu(t)dt.$$
We pick $T_n\uparrow+\infty$ such that
$$\left|\mu(\{x\})-\frac 1{2T_n}\int_{-T_n}^{T_n}e^{-itx}\widehat\mu(t)dt\right|\leq \frac 1n.$$
We can assume $T_n\geq n^2$. Removing $[-n,n]$, and using intermediate value theorem ($\widehat \mu$ is continuous), we can write
$$\mu(\{x\})=\frac{\widehat\mu(s_n)+\widehat\mu(t_n)}2,$$
where $s_n\to +\infty$ and $t_n\to -\infty$. This gives the result.