Why is the capacitor in this circuit not becoming an open circuit?
For a capacitor, Q = CV and, if we differentiate Q with respect to time we get current so: -

$$\dfrac{dQ}{dt} = I = C\dfrac{dV}{dt} $$

So, if \$I\$ is a constant current source then \$\dfrac{dV}{dt}\$ will be constant and that means that voltage will continue to rise linearly with time until your simulation runs out of numerical range or, in the real world, your "practical" current source is unable to generate the terminal voltage required to sustain a constant current.


  I thought that if we use a DC voltage source instead of a DC current
  source, this circuit would become open circuit after a while. Because
  the voltage on the capacitor is equal to the voltage source.


Correct.


  I've applied it the same way and thought about the case as DC current
  source. I think the result should be the same open circuit result.


Incorrect, for the reasons given above.