RC Circuit Charging and dischargin
Here is another way to look at the problem:

The capacitors are initially uncharged and they get charged instantly to same voltage 2.5V (since both have same capacitance) when the pulse is applied. Now that there is 2.5V across C1 (and hence across R), the resistor demands a current of 2.5mA at the instant.

Now the only path for the current is through C0. This charges up the capacitor C0. This is to say that capacitor C1 is discharging since the voltage across C0 and C1 must add up to 5V. 

You can also use Thevenin's theorem to gain proper insight into the circuits and obtain quicker solutions.

It is however easier to obtain the steady state values of voltages and currents by simplifying the circuit using complex impedance. If the circuit is of first-order (which usually is the case when problems of this kind are given), you would know that the step response would be exponential Here is how you do it:

The effective impedance of parallel combination of resistor R and 1/sC1 (impedance of capacitor in laplace domain) is:

$$
Z1 = \frac{R*(\frac{1}{sC1})}{R+\frac{1}{sC1}} = \frac{R}{1+sRC1}
$$
Z2 = 1/sC0

The steady state voltage across C1 would be:

$$
Vc1 =\frac{ VinZ1}{Z1+Z2}
$$
Since s=0 for DC voltages,
steady state impedance Z2 is infinity and Z1=R.

$$
Vc1 =0 
$$

$$
Vc0 = \frac{ VinZ2}{Z1+Z2} = \frac{Vin}{\frac{Z1}{Z2}+1} = 5V
$$
Once you know the steady state value, you can plot the required circuit parameters if you know the initial values. 

Instead of using vpulse for plotting circuit transients, you can use a DC source(vdc) and  set the initial voltage of capacitor (or initial current through inductor) to zero and run the simulation. If you don't, the simulator will show only the steady state value.