This is a valid method of deciding whether a program hangs. Another very simple approach is searching for global state recurrences (if your program goes back to exact same state at least once, and it is deterministic, then it never halts).

Note this has no bearing on the Halting problem though (famously proven impossible). You cannot decide whether an arbitrary very complicated program halts -- it may do all sorts of weird stuff that doesn't involve any of the previous behavior (having necessary subfunctions never terminate or global recurrence) -- it could try counting up to infinity until a certain complicated binary function of the number is true, then you have to decide the truthiness of this arbitrary function. In general you face a self-referential problem: imagine you try deciding whether a program of size N halts if it runs for too long. You'd need a program (say of size K) that produces a function f(N) of maximum running time for programs of size up to N -- then you could just check whether a program runs for more than f(N) steps. But by definition your program cannot run for more than f(K) steps, thus it cannot output a number greater than 2^(f(K)). Thus your program can only decide finitely many instances, and is not an algorithm (instead it is a sort of compressed look up table).

The proof of the Halting problem undecidability is more general, but uses similar self-referential arguments.

I like to think of it this way: an algorithm is really a compressor function. It compressed an infinite set of (input->output) relations into a finite program. But some infinite sets are inherently incompressible -- think a string of random numbers -- and in particular the set of all programs. Incompressible (actually finitely compressible) sets must exist because otherwise you would have infinitely many different sets that are output of finitely many programs, which is obviously impossible (this is the pigeonhole principle).