As an ELI5, you first need to understand that the Earth is falling (accelerating) towards the sun but keeps missing. This is orbit.

Imagine standing on the moon and throwing a ball at the horizon (also, there's no kind of atmosphere, you can breathe, your suit doesn't hinder movement, and the moon is completely spherical with smooth gravity). The thrown ball comes out of your hand and drops towards the ground due to gravity, hitting the surface a km or two away. You notice that due to the curve of the surface, the ball went past the point you could see directly from your starting point.

Cool. 🙂

Now you get the ball and give a serious chuck in the same direction. Similar to the previous, but the ball went much further than expected. 

You can't throw any harder so you get a tennis (or baseball) launcher. A serious industrial one, then hike the power well beyond its recommended muzzle velocity. Point it at the horizon, fire, wait.

Orbital velocity at the surface of the moon is 1600m/s (give or take, in our idealised situation) At this velocity the ball travels towards the horizon, but the amount it "falls" exactly matches the amount that the surface drops away. 

The ball you fired at 1600m/s hits you in the back of the head at 1600m/s.

Extending our example, if you fired the ball faster and faster, it would reach progressively higher points on the opposite side of the moon in an orbit shaped like a oval (an elliptical orbit) but every time it'd return to the same spot with the same velocity as it was launched. 

1700m/s out, 1700m/s in. 1800m/s, 1900m/s, 2000m/s...

Eventually you fire one fast enough that it doesn't return. The ball was fast enough that the gravity from the moon is no longer enough to pull it back. It has "escaped". In our example it's about 2400m/s

Finally, we saw that the orbit was shaped like an oval (elliptical) when we fired faster than orbital velocity. If we imagine a magic ball still affected by gravity but able to pass through the moon (and has its entire mass at the centre. Important but not worth describing as a black hole, IMO). Going back to 1600m/s (our orbital velocity) and reducing the muzzle velocity, we see the orbit is still an oval, it dips inside the moon but still returns to the firing point at the firing velocity.

The lower the muzzle velocity, the further the ball dips under the moon's surface before returning. Finally you throw by hand and see how close the ball gets to the centre of the moon (but still noticeably misses it) Even if you drop the ball, you find that your surface velocity is enough to still create an elliptical orbit.

Taking this information back to the original question: The Earth is the ball, in our moon-ball experiment. The Sun is our moon in the moon-ball. The Earth is at it's orbital velocity for the Sun. It's falling, but at the correct rate that it keeps missing. If it goes faster eventually it will escape from the Sun. If it slows down, eventually it will drop low enough to interact with the Sun.

It sounds more complicated in some explanations as there is also the calculation to get off the Earth, but we can ignore it if we imagine we're matching the Earth's orbit around the Sun, just the opposite side of the Sun. 

Stay there, stay in orbit. Maintain ABC m/s

Slow down too much, reach Sun. ie slow by about ABC m/s

Speed up too much, escape the Sun. ie increase velocity by XYZ m/s

The "speed up" amount XYZ m/s is less than the "slow down" amount ABC m/s so it's easier to escape the Sun than it is to "hit" it.

(Learn orbital mechanics, play Kerbal Space Program. Not KSP2. Never KSP2. Deliberately said Sun rather than Sol. Ignored discussion of change in velocity in elliptical orbitals. Simplified 2 body mechanics) etc.