Find an integer n such that:
|n| + |n - 1| + |n - 2| = 3
The way I have solved for the above was honestly eyeballing and seeing the absolute value of |-1|+|-2| = 3. Is there a better way to prove this? I'm worried about when the problems become more complex
Eyeballing this is okay, though there is another solution in addition to the one you found, should you have been asked to find all solutions. If you wanted to approach it more systematically to find all solutions, you could think about it in the following way, which is how it can be useful to work with equations involving absolute values. That is, $|A|=\pm A$.
An absolute value is either equal to what’s inside it or is the negative of what’s inside it. With three absolute values in this problem, you might first think there are up to eight different possibilities. On the surface, $|n| +|n-1|+|n-2| = {\color{red}\pm} n + {\color{red}\pm} (n-1) + {\color{red}\pm} (n-2)$, with eight possible combinations of plus signs and minus signs. But because the numbers $n$, $n-1$, and $n-2$ are consecutive integers, the three signs (in red above) can only be $(+,+,+)$ (if $n\ge2$), $(+,+,-)$ (if $n=1$), $(+,-,-)$ (if $n=0$), or $(-,-,-)$ (if $n\le-1$). (If one of the three terms is zero, two different sign sequences happen, but the cases I listed will catch every possibility.)
You can then solve for $n$ in each of these four cases. For example, in the $n\ge2$ case ($(+,+,+)$), your equation is $n + (n-1) + (n-2) = 3$, or $3n-3=3$, which leads a different solution than you found. To solve one more case, if $n\le-1$, the equation is $-n + -(n-1) + -(n-2) = 3$, or $-3n+3=3$, which looks like it leads to $n=0$ (which you found), though since this is the case where $n\le-1$ and $n=0\not\le1$, it doesn’t fit in this case and you’ll get the $n=0$ solution from a different one of the four cases.