Diode can be ideally treat as a switch. For the sake of understanding, please consider the following two circuits
simulate this circuit – Schematic created using CircuitLab
In left one, positive terminal of battery is connected to anode of diode and negative terminal of battery is connected to cathode of diode. Hence Anode relatively in higher potential than cathode. So when the applied voltage greater than barrier voltage of diode, diode is said to be forward biased.(closed condition) and hence a current flows through the circuit in the arrow direction.
In right one, negative terminal of battery is connected to anode and positive terminal is connected to cathode. So anode is relatively in lower potential than cathode. This makes diode in reverse bias mode, leaving the diode always in open circuit mode. So no current flow through the circuit.
Now consider our case. Here capacitor acts like a battery in negative half cycle.
simulate this circuit
So positive side of capacitor connected to cathode and negative side of capacitor is connected to anode. It is similar to the open circuit diagram (right side circuit in the first section). Since the anode is relatively lower potential than cathode, it is reverse biased mode. Hence no current flow through diode. Current will flow through load resistor only, as indicated by arrows.
Hopes this will make a good understanding. Thanks.