Borel Measures: Atoms vs. Point Masses Let a measure be $\mu:\Sigma\to\mathbb{R}_+$.
Call a measurable $A\subset\Sigma$ an atom if:
$$\mu(A)>0:\quad\mu(E)<\mu(A)\implies\mu(E)=0\quad(E\subseteq A)$$
and a singleton $\{a\}\in\Sigma$ a point mass if:
$$\mu(\{a\})>0$$
First of all, for singletons the notion of atoms and point masses coincide.
Next, for sigma-finite spaces the defintion of an atom is equivalent to:
$$\mu(A)>0:\quad\mu(A\cap E)=0\lor\mu(A\cap E^c)=0$$
(For more details see: Measure Atoms: Definition?)
That is every atom splits into smaller atoms, so one could try to apply Zorn's lemma here.
Now, given a sigma-finite Borel measure $\lambda:\mathcal{B}(\Omega)\to\mathbb{R}_+$.
Does every atom arise from a point mass?
 A: The example you give actually is a $\sigma$-finite Borel measure.  Equip $[0,1]$ with the cofinite topology (in which a set is open iff it is either empty or its complement is finite).  Then your $\Sigma$ is the Borel $\sigma$-algebra of the cofinite topology (it is a nice exercise to verify this).
However, there is the following result:
Proposition.  Let $(X,d)$ be a separable metric space, $\Sigma$ its Borel $\sigma$-algebra, and $\mu$ a $\sigma$-finite measure on $\Sigma$.  Then each atom of $\mu$ is the union of a point mass and a null set.
Proof.  Let $C$ be a countable dense subset of $X$.  For each integer $k\in\mathbb{N}$, we have $\bigcup_{x \in C} B(x, 1/k) = X$.  Thus $\bigcup_{x \in C} (A \cap B(x,1/k)) = A$.  So by countable additivity, there exists $x_k \in C$ such that $\mu(A \cap B(x_k, 1/k)) > 0$.  Since $A$ is an atom, $\mu(A \setminus B(x_k, 1/k)) = 0$.  Let $S = \bigcap_k B(x_k, 1/k)$.
Since for each $k$, $S$ is contained in a ball of radius $1/k$, $S$ contains at most one point.
On the other hand, by De Morgan's law and countable additivity,
$$\mu(A \setminus S) = \mu\left(\bigcup_k A \setminus B(x_k, 1/k)\right) = 0.$$
Since $\mu(A\cap S)=\mu(A) > 0$, $A\cap S$ is not empty, so $A\cap S$ is a singleton.
Hence $A\cap S$ is a point mass and $A \setminus S$ a null set.  $\Box$
So in this case, effectively the only atoms are point masses.
Note that we did not need to assume $X$ was complete.
For non-separable metric spaces, things are harder.  For uncountable discrete spaces (which are certainly metric), the question of whether there can be nontrivial  atoms is related to whether the cardinality of $X$ is a measurable cardinal, and such questions tend to be independent of the axioms of ZFC.  I asked a new question about it: Consistency strength of 0-1 valued Borel measures.
A: No, in general this is wrong!
Consider the Borel algebra generated by the cofinite topology:
$$\Sigma:=\{E\subseteq[0,1]:\#E\leq\aleph_0\lor\#E^c\leq\aleph_0\}=\sigma(\mathcal{T}_{cofinite})$$
(See the thread on: Cofinite Topology: Borel Algebra)
and the finite Borel measure:
$$\lambda(\#E\leq\aleph_0):=0$$
$$\lambda(\#E^c\leq\aleph_0):=1$$
Then the atoms are all the uncountables but no singleton is a point mass.
