Clarification on Constructing a Dense Borel Set that is Positive Measure but not Full Measure on Each Interval

I recently came across this exact same question asked here: Construction of a Borel set with positive but not full measure in each interval :

If $$\mu$$ denotes Lebesgue measure, how would one construct a Borel set $$A \subset \mathbb{R}$$ such that $$0 < \mu(A \cap I) < \mu(I)$$ for every interval $$I$$ in $$\mathbb{R}$$?

The answer with the highest vote seems to suggest the set $$A$$ constructed there is sufficient. I will copy-paste the answer for convinience:

Let $$\{r_n\}$$ be an enumeration of the rationals, let $$V_1$$ be a segment of finite length centered at $$r_1$$, and let $$V_n$$ be a segment of length $$m(V_{n-1})/3$$ centered at $$r_n$$. Set $$W_n=V_n-\bigcup_{k=1}^{\infty}V_{n+k},$$ and observe that $$$$m(W_n)\geq m(V_n)-\sum_{k=1}^{\infty}m(V_{n+k})=m(V_n)-m(V_n)\sum_{k=1}^{\infty}3^{-k}=\frac{m(V_n)}{2}.$$$$ In particular, $$m(W_n)>0$$. For each $$n$$, choose a Borel set $$A_n\subset W_n$$ with $$0. Finally, put $$A=\bigcup_{n=1}^{\infty}A_n$$. Because $$A_n\subset W_n$$ and the $$W_n$$ are disjoint, $$m(A\cap W_n)=m(A_n)$$. That is to say, $$0 for every $$n$$. But every interval contains a $$W_n$$, so $$A$$ meets the criteria, and has finite measure (specifically, $$m(A)\leq\sum_n m(V_n)=2 m(V_1)<\infty$$). In particular, we must have for all interval $$I$$, $$0 < m(A \cap I) < m(I).$$

Now I can show why $$0 < m(A \cap I)$$ is true: Since every interval $$I$$ must contain some constructed rational centered interval $$V_n$$, we have as $$A_n \subseteq W_n \subseteq V_n$$ for all $$n \in \mathbf{N}$$, $$m(I \cap A) \geq m(V_n \cap A) = \sum_{i = 1} ^\infty m(V_n \cap A_i) \geq m(V_n \cap A_n) = m(A_n) > 0.$$ But I can not seem to prove $$m(I \cap A) < m(I).$$ Any suggestions as to how to proceed?

Let $$I$$ be a nonempty interval. Then it contains $$W_n$$ for some integer $$n$$, and \begin{align} m(A\cap I) &=m(A\cap W_n)+m(A\cap(I\backslash W_n))\\ &