Is there a direct proof of the following? I have been warned by my Lecture as well as several other sources that while proof by contradiction  is useful and is certainly  needed in some cases, it is often overused.
In a effort to learn, I was wondering if a direct of the proof of following  statement 
was possible:
"If $\{B_i\}_{i \in I } $ is  a pairwise disjoint indexed family of sets then $\{B_i\}_{i \in I } $ is  disjoint."
As opposed to  the following:
Suppose  $\{B_i\}_{i \in I } $ is  a pairwise disjoint indexed family of sets, but $\{B_i\}_{i \in I } $ is not disjointed
then by  definition :
$$\bigcap_{i \in I} B_i  \neq  \varnothing  $$
i.e. the exists a $x$ such that $x \in B_i$ for all $i \in I $.
However  since $\{B_i\}_{i \in I } $  is pairwise disjoint
$$B_i  \cap  Bj  = \varnothing \text{  where }  i,j  \in  I  \text { and } i \neq j $$
Hence we reach a contradiction.
or something in a similar vein ( I am not happy with the above proof by contradiction as I think it is not as concise  and precise as I should be)
Note: Here are the definitions I am working with:
An indexed family of sets $\{B_i\}_{i∈I}$ is said to be disjoint if $ \bigcap_{i \in I}  B_i  = \varnothing $.
The family is said to be pairwise disjoint if $B_i \cap B_j  =  \varnothing $
 whenever $i  \neq j$
 A: You could certainly use the contrapositive instead of contradiction: prove that if the family $\{B_i\}_{i\in I}$ is not disjoint, then it is not pairwise disjoint.  The proof would be almost identical with what you have done, but you could dispense with your initial assumption that the family is pairwise disjoint.
This may not be quite what you would regard as a direct proof, but it does get rid of the use of contradiction.
You could even attempt to eliminate the contrapositive as follows.  Suppose that the family is pairwise disjoint.  Then by definition, no $x$ is in more than one of the sets $B_i$.  Therefore no $x$ is in all the $B_i$, and so the family is disjoint.
If you accept that the statement I have put in italics is sufficiently obvious be be used without further explanation, then this is a direct proof.  If you consider it needs further explanation, then I suspect you may be forced back to using the contrapositive.
A: For the index set I let $j,k$ be specific members of I. Then,
$\bigcap_{i \in I} B_i   = B_j \bigcap B_k \bigcap_{i \in I'} B_i$ where $ I' = I $\ {j, k} (assuming that |I| > 2).
Since intersection is associative then $\bigcap_{i \in I} B_i   = (B_j \bigcap B_k )   \bigcap_{i \in I'} B_i$ = $\emptyset \bigcap_{i \in I'} B_i$ = $\emptyset$
