No, there are non-atomic measures whose support is a set of measure zero.
These are called singular continuous measures and they are not absolutely continuous with respect to the Lebesgue measure.
For example, the measure corresponding to the devil staircase function has as it's support the Cantor set, but the measure of every point is zero.
Let $F$ be the devil staircase function on the unit interval. Then we can define a Borel measure on the unit interval which is singular with respect to the Lebesgue measure and has no atoms as follows: for $E\subset[0,1]$ set
$$\mu_*(E)=\inf\sum_{j=1}^\infty F(b_j)-F(a_j)\tag{1}$$
where the infimum goes over all coverings of $E$ with intervals $(a_j,b_j)$, $j=1,2,\dots$.
This is the Riemann-Stieltjes outer measure associated to $F$. Go ahead and prove that it really is an outer measure. Then we apply Caratheodory's extension theorem (as in the construction of the Lebesgue measure) to obtain a corresponding Borel measure $\mu$ (such that $\mu_*(E)=\mu(E)$ for measurable $E$).
Note that $\mu((a,b))=F(b)-F(a)$. Since $F$ is continuous, it really follows that $\mu(\{x\})=F(x)-F(x)=0$. On the other hand, the measure is not absolutely continuous, because $\mu(E)=\mu(C\cap E)$ for every measurable $E$, where $C$ is the Cantor set. Since $\lambda(C)=0$ ($\lambda$ being the Lebesgue measure) we see that $\mu$ is singular with respect to the Lebesgue measure.
Since you wanted a measure on the real line just extend this measure on the unit interval by $0$ outside of $[0,1]$.
This construction works with any increasing function and it provides (after proper normalization) a 1:1 correspondence between increasing function and positive Borel measures (see Stein, Shakarchi Real Analysis).
Note:
Every positive Borel measure $\mu$ (say on the unit interval, it also works on $\mathbb{R}$ but there you need $\sigma$-finiteness as additional assumption) can be uniquely decomposed as
$$\mu=\mu_a+\mu_{sc}+\mu_{pp}$$
where $\mu_a$ is absolutely continuous, $\mu_{sc}$ is a singular continuous measure (i.e. singular and no atoms) and $\mu_{pp}$ is a pure point measure, i.e. the support consists entirely of atoms (all with respect to Lebesgue measure, but also other measures can be used).
Look up the Radon-Nikodym theorem in Rudin's Real and Complex Analysis for a reference.
Let $\alpha_n = \lambda(B_1(0))$.
Since $\lambda(E) = \alpha_n$ you have $\lambda(E \setminus F) + \lambda(E \cap F) = \alpha_n$.
However $\lambda(E \setminus F) \le \lambda(E \triangle F) \le \lambda(B_{\frac 12}(x)) = \left( \frac 12 \right)^n \alpha_n$.
You can arrange these to find $\lambda(E \cap F) \ge \left( 1 - \left( \frac 12 \right)^n \right) \alpha_n$ giving you what you need if $n \ge 2$.
Best Answer
For the first question:
Let $A = \bigcup_{n \in \mathbb N} \{a_n \} \subset \mathbb R$. The Lebesgue measure of a point is zero: by construction of the Lebesgue measure, $\lambda [a,b] = b - a$. For the one element set $\{ a \} = [a,a]$ we have $\lambda [a,a] = a-a =0$.
Since the Lebesgue measure is additive, we have $$ \lambda A = \sum \lambda \{a_n\} = \sum 0 = 0$$