All the voltage measurements are differential. That is we measure Vin+ to Vin-, Vout+ to Vout-.
If you prefer, you can look at it this way.
As no current flows round the circuit, the voltage across the capacitor does not change, it stays at zero. The Vin graph you posted shows the Vin- voltage going 1V 'down' with respect to the Vin+ terminal. As the voltage across the capacitor stays at 0V, the Vout- terminal also goes 1V down with respect to the Vout+ terminal, as shown in the Vout graph. There's no need to worry about the voltages on the capacitor terminals 'moving' at all.
Of course, as there's no other voltage reference drawn, no ground terminal, it is only differential voltages that are meaningful. So the question of do the capacitor terminals go up in voltage is meaningless, their voltage can only be referenced to other terminals in the circuit.