Suppose $m(A)>0$. Note that $m(A)\leq 1$ since $A\subset [0,1]$. Let $\epsilon=m(A)/4$ which is a finite positive number. For this chosen $\epsilon>0$, we can choose a countable union of open interval $(a_i,b_i)$, $i\geq 1$ such that $A\subset \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$ and $\tag{1}-\epsilon+\sum_{i=1}^\infty(b_i-a_i)\leq m(A)\leq \sum_{i=1}^\infty(b_i-a_i).$ Then $\tag{2}m(A)=m(A\cap\bigcup_{i=1}^\infty(a_i,b_i)) =m(\bigcup_{i=1}^\infty A\cap(a_i,b_i))\leq\sum_{i=1}^\infty m( A\cap(a_i,b_i))$ where the first equality follows from $A\subset \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$, and the last inequality follows from the subadditivity of measure. On the other hand, $\tag{3}\sum_{i=1}^\infty m( A\cap(a_i,b_i))\leq\sum_{i=1}^\infty m( A\cap[a_i,b_i]) \leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}$ where the last inequality follows from assumption.
Now combining $(1)$, $(2)$ and $(3)$, we have $-\epsilon+\sum_{i=1}^\infty(b_i-a_i)\leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}$ which implies that (since $\sum_{i=1}^\infty(b_i-a_i)<\infty$ by $(1)$, we can do the subtraction) $\sum_{i=1}^\infty\frac{b_i-a_i}{2}\leq \epsilon=\frac{m(A)}{4}$ which contradicts $(1)$ when $m(A)>0$.
Therefore, we must have $m(A)=0$.