I don't understand why escape velocity is necessary I have read multiple explanations of escape velocity, including that on Wikipedia, and I don't understand it.
If I launch a rocket from the surface of the Earth towards the sun with just enough force to overcome gravity, then the rocket will slowly move away from the Earth and we see this during conventional rocket launches.
Let's imagine I then use slightly excessive force until the rocket reaches 50 miles per hour and then I cut back thrust to just counterbalance the force of gravity. Then my rocket will continue moving at 50 mph toward the sun. I don't see any reason why I can't just continue running the rocket at the same velocity and keep pointing it towards the sun. The rocket will never orbit earth (by "orbit" I mean go around it). It will just go towards the sun at 50 mph until it eventually reaches the sun. There seems to be no need whatsoever to ever go escape velocity (25,000 mph).
Answer: just to clarify the answers from below and the other linked question... Escape velocity is not necessary to leave the earth, unless the object has no thrust or other means of propulsion. In other words, if you throw a baseball, it has to go escape velocity to leave the earth, but if you have a spaceship with engines, then you leave the earth at any speed you want as long as you have the fuel necessary.
 A: As per other answers, your operational example simply doesn't correspond to the >>definition<< of "escape velocity". An operational example that does correspond to the definition is the cannon in Jules Verne's classic sci fi story https://en.wikipedia.org/wiki/From_the_Earth_to_the_Moon
So what you're suggesting can absolutely be done exactly as you say, and it will work exactly as you say. But it has nothing to do with "escape velocity". That, instead, would be the minimum speed a cannoball has to be fired with to just escape the Earth (ignoring atmospheric resistance), with >>no further forces<< after it's initially fired. That's just the definition of the term.
A: Escape velocity is the velocity an object needs to escape the gravitational influence of a body if it is in free fall, i.e. no force other than gravity acts on it. Your rocket is not in free fall since it is using its thruster to maintain a constant velocity so the notion of "escape velocity" does not apply to it.
A: Escape velocity is necessary if you turn all your rockets off.  Without any rocket thrust, and if you are going below escape velocity, you will go into orbit or crash into the object you are trying to escape.  If you keep your rockets on, you don't need escape velocity. 
To put this into practical terms,  during the Apollo lunar missions,  the rockets were off nearly all of the time.  The burn called "translunar injection"  gave the spacecraft enough velocity to get out of earth's gravitational influence into the moon's.  Another burn was needed when the spacecraft arrived in the neighborhood of the moon.  This burn was to put it into lunar orbit.  There was another burn needed to escape the lunar pull, and put it back on a trajectory to earth.  The were a few minor burns for mid course correction.  And there was the great big burn, at launch time.  
The lunar exploration module had to make a few more burns,  to land on the moon, to return to lunar orbit,  and to link up with the mother ship.
Other than that, the trajectory of the craft was determined by gravity and inertia  (momentum).  
A: If you're in a car at highway speeds and jump out in any direction, are you still going at highway speeds?
If you shoot a ball at 50mph with a cannon out of the back of a car that's driving 50mph it would stand still:
https://www.youtube.com/watch?v=BLuI118nhzc
So imagine Earth is the car where you're jumping out of, and the Earth is traveling around the Sun at about 30km per second. You'd still need to add 19km/s to the "escape velocity" in the opposite way before you come to a standstill and fall down towards the sun. 19km/s is a lot of fuel!
A: 
If I launch a rocket from the surface of the Earth towards the sun with just enough force to overcome gravity, then the rocket will slowly move away from the earth and we see this during conventional rocket launches.

This is not what happens in actual spaceflight. The actual rockets work for a short time and after that, the spacecraft is moving by inertia. And they don't really work against the Earth's gravity - the vertical launch purpose is to shoot the rocket to the high altitude where the atmosphere is thin. Then the rockets turn and accelerate horizontally to gain enough velocity to get on the orbit or the desirable escape trajectory. What you describe would be extremely inefficient and no rocket exists to actually do that in real life.

To understand why no rocket can reach far this way let's do a quick calculation of the amount of fuel required. Let's assume that going at a constant speed 50 mph (80 km/h) you want to reach an 80 km altitude (the altitude one needs to be awarded by the astronaut wings in US). At that altitude, the gravity acceleration $g$ is almost the same as on the ground. That's why we will assume it to be constant. Then you rocket fighting this acceleration for a 1 hour should have so much fuel that if it were in an empty space without any gravitating body it would speed itself to the velocity equal $\Delta v= 1\ \mathrm h\cdot g$. The Tsiolkovsky equation relates this speed to the ratio of the mass of the fueled rocket $m_0$ to its final mass $m_\mathrm f$.
$$\frac{m_\mathrm f}{m_0}=\exp\left[\frac{\Delta v}{g I_\mathrm{sp}}\right]=\exp\left[\frac{1\ \mathrm h}{I_\mathrm{sp}}\right]$$
where $I_\mathrm{sp}$ is a so-called specific impulse depending on the type of the rocket. For the idealized LH2-LOX rocket $I_\mathrm{sp}=450\ \mathrm s$. This means that for such rocket $\frac{m_\mathrm f}{m_0}=\mathrm e^{8}\simeq 2980$. I.e. to elevate 1 ton just to this altitude this way you need the same amount of fuel as the mass of the whole Saturn V rocket. And this computation is idealized i.e. all rocket engines, the supporting structure, fuel tanks etc are included into this 1 ton. If we raise the altitude the mass ratio grows exponentially i.e. you need $\simeq 10^{17}$ tons of fuel just to elevate 1 ton to the altitude of the ISS.
