Question: ![enter image description here](https://i.stack.imgur.com/6EXpa.png)

Why does the object not go inward, into the circle if the acceleration is inward? I think its because the velocity to outward? So they sort of cancel each other out? But if the speed is kept constant and the acceleration is inward, won't the object eventially move inward? 

Also, what keeps the direction of acceleration always point inward?

Answer: 
I suspect Luboš's answer may be a bit complex for you so I'll attempt a simpler explanation (if I'm wrong just ignore this answer).

When you say "I think its because the velocity to outward" you're getting close. To show what's going on I've zoomed in on the diagram you posted in your question.

![Acceleration](https://i.stack.imgur.com/N6eQT.gif)

Let's start at the moment where the velocity of the object is $u$. The thing you need to remember is that *velocity* is a vector so it has both magnitude and direction. Acceleration may be a change in the magnitude or a change in direction or both. In the case of circular motion it's the direction of the velocity that is changing not it's magnitude.

To see this look at the vector $v$, which is the velocity of the particle after some short time. So the velocity has changed from $u$ to $v$. To get the size of the change we need to find what vector has to be added to $u$ to give $v$. At the top of the diagram I've put the two vectors $u$ and $v$ with their starting point together, so you can see that the vector needed to change $u$ to $v$ is the small vector $a$, and you can see it points downwards. If I move $a$ to the starting point of $u$ on the circle you can see that $a$ points towards the centre of the circle.

The vector $a$ is of course the acceleration (well it's the acceleration times the short time interval between $u$ and $v$) and that's why the acceleration points towards the centre of the circle.

If you didn't have any acceleration the object would move in a straight line so it would just move away at a tangent to the circle. The effect of the acceleration $a$ is to bend the trajectory of the object so it stays on the circle.

Finally, you ask "what keeps the direction of acceleration always point inward". You've mixed up cause and effect. It isn't that there's some magical property of circular motion that causes the acceleration to be central. It's that if in some system the acceleration is always central the object will move in a circle (or strictly speaking in a conic section, but let's not complicate matters).

For example suppose I tie a stone to a string and start whirling it round my head. The force causing the acceleration is provided by the string, and obviously the string always runs from the stone to the centre because my hand is at the centre. It's because the force caused by the string is always central that the stone moves in a circle.
No, a uniform circular motion has both constant speed and constant acceleration and it will continue indefinitely, never going inwards.

If there were no inward acceleration, the object would move along the straight line in the direction of $\vec v$, and therefore away from the circle. To keep the object on the circle, there has to be a radial attractive force – the centripetal force – whose magnitude must be $F=ma=mv^2/r$.

We may also describe the situation from the viewpoint of the frame rotating together with the object. In that frame, there is a centrifugal, outward force $mv^2/r$, which tries to "repel" the object from the origin. This centrifugal (fictitious) force must be compensated by a real, centripetal force if the object is supposed to stay at a fixed distance from the origin.

Those statements may be easily derived from a particular coordinate parameterization of the motion. The coordinates are
$$ x(t) = r\cdot \sin(\omega t), \quad y(t) = r\cdot \cos(\omega t)$$
The velocities are derivatives with respect to $t$
$$ v\_x(t) = r\omega \cdot \cos(\omega t), \quad v\_y(t) = -r\omega\cdot \sin(\omega t)$$
and the accelerations are derivatives of the velocities
$$ a\_x(t) = -r\omega^2\cdot \sin(\omega t), \quad a\_y(t) = -r\omega^2\cdot \cos(\omega t)$$
and they are proportional to $(x,y)$ themselves, so the force is inward and radial. Note that $r\omega=v$ so $r\omega^2=v^2/r$.