Is it possible to find a set of 5 integers, such that the 10 sums each have a different last digit? Full question: 

If we are given 5 integers, there are $\binom{5}{2} = 10$ different ways to find the sum of 2 of the integers.
  Is it possible to find a set of 5 integers, such that the 10 sums each have a different last digit?

Am I trying to prove that 5 integers can be found, such that the 10 sums each have a different last digit, or am I being asked to find 5 such integers?
 A: Let $o$ be the number of odd integers in the set. Then you can form $o(5-o)\ne5$ odd sums. Therefore a set of the desired kind does not exist.
A: Idea 1) you have only $10$ possible sums and you need ten different results. 
Idea 2) Let's look at parity: at least one number is odd and at least one number is even.


*

*$1$ odd and $4$ even: no, because this gives you only $4$ odd sums, and you need five.

*$2$ are odd and $3$ are even: no, because this gives you $6$ sums with odd result. So at most $4$ sums have even results, while you need at least five of them.

*$3$ are odd and $2$ are even: no, because, again, you have $6$ odd sums.

*$4$ are odd and one is even: no, because you have only $4$ odd sums.
A precisation: For the second and third case, one must be careful to what is an odd result and what is an non-ordered pair $\{ a,b\}$ such that $a+b$ is odd. In total, the number of non-ordered pairs at your disposal is $10$. If you have (say) $2$ odd numbers and $3$ even numbers, the number of non-ordered pairs $\{ a,b\}$ such that $a+b$ is odd is exactly $6$. Which leaves at most $4$ non-ordered pairs $\{ c,d\}$ with $c+d$ even, hence (at most) four even possible sums. This is where idea $(1)$ comes into play.
