How does a capacitor blocks low frequency signal? Please explain it without the impedance formula?
When you talk about 'block' then a simple proper context is that the capacitor is in series with some kind of load (let's assume a resistor), and that there is a voltage input to this with some frequency.

If the frequency is low enough (for a given capacitance and load values), then as the voltage follows its low frequency sinusoidal curve, the capacitor will have plenty of time to get charged/discharged and will follow this voltage closely.  So there is basically not much voltage 'left' for the load - the capacitor is blocking it.  

Put another way, since the capacitor charges/discharges with little current (because current is proportional to the derivative of its voltage in time), and the drop on the resistive load is proportional to current, then not much voltage will drop on the load - it will be attenuated and close to zero.

As you increase the frequency, the capacitor will not have much time to fully charge/discharge, so its voltage will have an attenuated sinusoidal shape (that is out of phase with respect to the input power supply).  The rest dropped on the load.  [By the way, the current that the capacitor has available to get charged/discharged is limited by the load resistor itself, since it is in series with it].

If the frequency is high enough, the capacitor will barely charge/discharge, and most of the input voltage will be seen at the load, as if the capacitor didn't exist and was replaced with a wire.  It basically let's through high frequencies.