In general, if $A_1, A_2,\ldots, A_n$ are mutually disjoint events, then
$$
P\Bigl(\,\bigcup\limits_{i=1}^n A_i\,\Bigr ) =\sum_{i=1}^n P(A_i).
$$
But (1) does not hold for arbitrary unions. One strategy to find a formula for the probability of a union of arbitrary events is to write the given union as a (different) union of mutually disjoint events.
From this, it follows that your initial statement, "This is the probability of sample points belonging to exactly 1 event, exactly 2 events, ...,exactly $n$ events." , is correct.
$P\Bigl(\,\bigcup\limits_{i=1}^n A_i\,\Bigr )$ is the sum of the probabilities of the events $B_i$ where $B_i$ is the event that exactly $i$ of the events $A_j$ occurs.
You have, noting that the $B_i$ are mutually disjoint:
$$
P\Bigl(\,\bigcup\limits_{i=1}^n A_i\,\Bigr ) =\sum_{i=1}^n P(B_i).
$$
But, the subsequent formula you have is incorrect. In fact, the event $B_1$ is somewhat complicated. If exactly one event $A_i$ occurs, then either $A_1$ only occurs, or $A_2$ only occurs, ... .
So
$$
B_1= \bigcup_{j=1}^n \Bigl(\,A_j\cap ({\textstyle\bigcup\limits_{i\ne j }}\, A_i^C\,)\,\Bigr).
$$
Trying to write down descriptions of the other $B_i$ might lead you to suspect this particular method is more trouble than its worth (of course, it may prove useful in certain problems).
The formula in Dilip's comment is a nice way to "disjointify" the union $\cup A_i$. If you draw a Venn diagram for the case $n=2$, it is easy to convince yourself that
$A_1\cup A_2$ can be written as a union of pairwise disjoint events:
$$
A_1\cup A_2 = A_1 \cup (A_1^c\cap A_2).
$$
So you can write
$$
P(A_1\cup A_2) =P( A_1)+ P (A_1^c\cap A_2).
$$
For the case $n=3$, $A_1\cup A_2\cup A_3$ can be written as a union of pairwise disjoint events:
$$
A_1\cup A_2\cup A_3 = A_1 \cup(A_1^c\cap A_2)\cup (A_1^c\cap A_2^c\cap A_3).
$$
So you can write
$$
P(A_1\cup A_2\cup A_3) =P( A_1)+P(A_1^c\cap A_2)+P (A_1^c\cap A_2^c\cap A_3).
$$
And, of course the general formula in Dilip's comment holds because the union of the $A_i$ is written as a union of pairwise disjoint sets.
The "add and subtract business" you mention seems like what is done in deriving the so-called Inclusion-Exclusion principle, as mentioned in gnometorule's answer. This gives a different method for evaluating the probability of a union (it is not exactly the "disjointification" method above).
As a warm up to the general formula, let's consider the formulas for $P(A\cup B)$ and $P(A\cup B\cup C)$.
Probability of the union of two events:
Let us find the probability of the union of two arbitrary events $A$ and $B$.
One might think $P(A\cup B)=P(A)+P(B)$; however, each of
$P(A)$ and $P(B)$ counts the probability of $A\cap B$. We thus have to subtract this probability from $P(A)+P(B)$ to obtain the correct formula:
$$
P(A\cup B)=P(A)+P(B)-P(A\cap B).
$$
Another way to derive the formula is to note that $A\cup B$ is the disjoint union of $A/B$, $B/A$, and $A\cap B$.
Now, since $A$ is the disjoint union of $A\cap B$ and $A/B$, it follows that $P(A)=P(A\cap B)+P(A/B)$; whence $P(A/B)= P(A)-P(A\cap B)$.
In a similar manner, one shows that $P(B/A)= P(B)-P(B\cap A)$.
It follows that
$$\eqalign{
P(A\cup B)&=P(A/B)+P(B/A)+P(A\cap B)\cr
&=\bigl( P(A)-P(A\cap B)\bigr)+\bigl(P(B)-P(B\cap A)\bigr)+P(A\cap B) \cr
&= P(A)+P(B)-P(A\cap B).
}
$$
Probability of the union of three events:
For three arbitrary events $A $, $B$, and $C$, to find $P(A \cup B\cup C)$, we first start with the guess
$$
P(A\cup B\cup C)=P(A )+P(B)+P(C).\tag{1}
$$
Now (1) counts each of $P(A \cap B)$, $P(A\cap C)$, and $P(B\cap C)$ twice. To make up for this,
we have to subtract
$$P(A \cap B)+P(A \cap C)+P(B\cap B).\tag{2}$$
But $P(A )+P(B)+P(C)$ counts $P(A \cap B \cap C)$ thrice, and in (2) we subtracted it thrice. So, we must add
$P(A )+P(B)+P(C)$ back in again. We have:
$$
P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).
$$
Looking at the Venn diagram may be helpful here:
More generally, for the events $A_1$, $A_2$, $\ldots\,$, $A_n$, we have the inclusion-exclusion principle:
$$\eqalign{
P\Bigl(\bigcup_{i=1}^n A_i\Bigr) = \sum_{i\le n} P(A_i) - &\sum_{i_1<i_2} P(A_{i_1}\cap A_{i_2})
+\sum_{i_1<i_2<i_3} P(A_{i_1}\cap A_{i_2}\cap A_{i_3}) - \cr
&\cdots+ (-1)^{n}\sum_{i_1<i_2<\cdots<i_{n-1} }
P(A_{i_1}\cap\cdots\cap A_{i_{n-1}} )\cr&\qquad\qquad + (-1)^{n+1}P(A_1\cap A_2\cap\cdots\cap A_n)}.
$$
As pointed out in the comments, this can be proved by induction.
Best Answer
We can write $(\bigcap_{i=1}^n A_i)^c = \bigcup_{i=1}^n (A_i^c)$. Then bound this in the same way as the first inequality and finish the proof by using that $Pr(A^c) = 1 - Pr(A)$.