Another elementary proof: Assume that $(x_n)$ is not weakly null. Then, passing to a subsequence, we get a functional $x^*$ such that $\inf_n |x^*(x_n)|>0$. By unconditionality, there are $c_1$, $c_2$ $c_3>0$ such that $$ \Vert \sum_n a_n x_n\Vert \ge c_1 \Vert \sum_n \epsilon_n a_n x_n\Vert \ge c_2 \vert \sum_n \epsilon_n a_n x^*(x_n) \vert= c_2 \sum_n \vert a_n x^*(x_n) \vert \ge c_3 \sum_n \vert a_n\vert, $$ where $(\epsilon_n)$ is a suitable choice of signs.

A more detailed version of the above proof: Assume that $(x_n)$ is not weakly null. Then, passing to a subsequence, we get a functional $x^*$ such that $\inf_n |x^*(x_n)|>0$. By unconditionality, there are $c_1$, $c_2$ $c_3>0$ such that $$ \Vert \sum_n a_n x_n\Vert \ge c_1 \Vert \sum_n \epsilon_n a_n x_n\Vert \ge c_2 \vert \sum_n \epsilon_n a_n x^*(x_n) \vert= c_2 \sum_n \vert a_n x^*(x_n) \vert \ge c_3 \sum_n \vert a_n\vert, $$ where $(\epsilon_n)$ is a suitable choice of signs.

Another elementary proof: Assume that $(x_n)$ is not weakly null. Then, passing to a subsequence, we get a functional $x^*$ such that $\inf_n |x^*(x_n)|>0$. By unconditionality, there are $c_1$, $c_2$ $c_3>0$ such that $$ \Vert \sum_n a_n x_n\Vert \ge c_1 \Vert \sum_n \epsilon_n a_n x_n\Vert \ge c_2 \vert \sum_n \epsilon_n a_n x^*(x_n) \vert \ge c_3 \sum_n \vert a_n\vert, $$ where $(\epsilon_n)$ is a suitable choice of signs.

Another elementary proof: Assume that $(x_n)$ is not weakly null. Then, passing to a subsequence, we get a functional $x^*$ such that $\inf_n |x^*(x_n)|>0$. By unconditionality, there are $c_1$, $c_2$ $c_3>0$ such that $$ \Vert \sum_n a_n x_n\Vert \ge c_1 \Vert \sum_n \epsilon_n a_n x_n\Vert \ge c_2 \vert \sum_n \epsilon_n a_n x^*(x_n) \vert= c_2 \sum_n \vert a_n x^*(x_n) \vert \ge c_3 \sum_n \vert a_n\vert, $$ where $(\epsilon_n)$ is a suitable choice of signs.