Here's an explanation adapted from
here. (I also skimmed Littlewood's
original paper, but it's not my field of expertise and it's in French, so it was a pretty rough skim.)
Imagine a bunch of wheels, each with an arrow pointing from its center to its edge, spinning at different, unrelated speeds. Eventually -- and it might take a very long time -- there will be a moment when all the arrows are coincidentally pointing in the same direction. It's worth taking a minute to think about why this is true for 2 or 3 wheels. Importantly, we know that this happens even if we don't know exactly how fast all the wheels are spinning -- we just won't be able to predict exactly when.
There is actually a formula which can be used to calculate the prime counting function precisely. It works by starting with the approximation mentioned in the main entry, the logarithmic integral function, and then adding an infinite series of smaller and smaller corrections to it. The biggest of these is always a negative number, which is why, at the outset, the prime counting function is smaller.
But you can think of the rest of these corrections as roughly described by wheels with arrows on them, which rotate as we move through the number line. For each correction, when its arrow is pointing up, it's a positive number; when the arrow is pointing down, it's a negative number; as the wheel spins, it oscillates between plus and minus. (The rates at which these wheels spin are a sequence of numbers just as mysterious as the prime numbers, so this formula isn't the most useful for direct computation, but it's a valuable change of perspective.)
So the essential principle of Littlewood's argument was the idea at the beginning of this comment: eventually, enough of these corrections are going to coincidentally line up pointing in the positive direction that they'll push the prime counting function above the li(x) approximation.
(The biggest problem with this analogy is that there are now infinitely many wheels, so we can't talk about them all lining up at once. The question is now a trickier one of getting
enough of them to line up. But the idea is similar.)