Proving NP-completeness intuition

There are problems that are still not known to be in $P$ or $NP$-hard, so there probably is no definite way to tell that. Moreover, it is possible that some problems are neither in $P$ or $NP$-hard, just in-between. Still I can share my own heuristits when deciding if something is in $P$ or $NP$-hard.

1. Problems in $P$ usually have regular structure and good locality--it is enough to know few local things and some small global state.
2. For problems in $P$ there are usually some bounds that restrict possible number of choices at a given moment disregarding previous decisions, e.g. in BFS algorithm in every moment the size of the queue is at most $n$, independent of which paths you have traversed or which vertexes have visited; similarly the sum of neighbors of all vertexes is bound by $m$, independent of the path by which you have reached them.
3. Problems that are $NP$-hard usually have a subset nature hidden somwhere, i.e. SAT have subsets of variables that are true, TSP has subsets of edges, graph-coloring algorithms have subsets of vertices with given color, Clique has subsets of vertices, etc.
4. Problems that are $NP$-hare are usually non-local--changing color of one vertex may possibly force majority of other colors to be changed, changing one SAT variable may force a lot of others to change their values.
5. Computing some numbers with arbitrary precision can be $NP$-hard; then the intuition is that in that number you could encode a complex problem.
6. There is a class of problems called $FPT$ (Fixed Parameter Tractable), often $NP$-hard problems would have some kernel that is hard, and then if that kernel can be somehow bounded, then the algorithm may be still polynomial; e.g. many graph problems that are $NP$-hard, can solved in $P$ for planar cases or graphs with bounded tree-width.
7. I could go one further, but the hints are less and less usable and I have to finish somewhere, so finally, it is really important what is the input to the algorithm--Knapsack problem is $NP$-complete, but there is a dynamic programming approach that is polynomial if the size of the knapsack is given in base $1$ (i.e. as a string of zeros of appropriate length).

Have fun with those!