Solution:

Expanding each number as the sum of the two previous numbers gives, for example,
$$1-2+3-5+8-13=1-(1+1)+(1+2)-(2+3)+(3+5)-(5+8)\text{.}$$
Now, after removing parentheses, this is a telescoping series: each term subtracts out with its neighbor, so what's left is the last term of this expanded sum, or the second-to-last term of the original sum.
$$=1-1-1+1+2-2-3+3+5-5-8=-8\text{.}$$
To make it clear that this pattern always works, it would be nice to write all of this in terms of $F_n$, the $n^{\text{th}}$ number in the list. When doing so, the previous expression becomes
$$F_1-F_2+F_3-\ldots \pm F_n=\left(F_1\right)-\left(F_0+F_1\right)+\left(F_1+F_2\right)-\ldots \pm\left(F_{n-2}+F_{n-1}\right)$$
and so this works for all $n$ because $F_0=F_1$, so every pair of consecutive terms in this long expression adds up to $0$, except for the $F_{n-1}$ at the end.

Pair up the terms. For our example above,
$$1-2+3-5+8-13=(1-2)+(3-5)+(8-13)=1+2+5,$$
which shows that after the initial $1$ we get every other term of the sequence. Then, we add, with $1+2=3$, and then $3+5=8$. To make it clear that this pattern always works, you will need to consider examples with an odd number of terms as well as showing that $F_1+F_2+F_4+\ldots+F_{2n}=F_{2n+1}$.
Other good arguments include a proof by mathematical induction, which supposes that this pattern works for the sum up to $F_n$ and then proves that it continues working for the sum up to $F_{n+1}$.