When current flows why diode not makes any shortage in it
There are few things that are not necessarily correct and it depends on what you mean.


  there are 2 wires going out from the source of power 1 is going to diode and 2nd is going to capacitor


One node is usually considered to be a reference node. All voltages are with respect to that node. This is usually indicated with a ground symbol, however, you're schematic does not have one, so it's implied that the the bottom of Vs is considered to be the reference node. However, its not necessary for that node to be the reference node, it just makes it easier to analyse the circuit.


  normally when we touch plus and minus 


If you connect a piece of wire or some very small impedance then you have a short. Otherwise its not a short. So if you use a meter to measure the voltage, it's not a short because the voltmeter will have a high impedance.

The title of your question does not reflect the body of the question, so I have only described the points you mentioned in the body of your question.

added based on comments

It's best not to describe a signal as plus and minus, because those terms are used to indicate polarity. Instead begin using the term source (the origin of the signal). 

A capacitor is a has an impedance that is dependent on frequency.

$$ Z_c = \frac{1}{j\omega C} $$

As frequency, \$\omega \$, increases, Zc will decrease. So at high enough frequencies the capacitor will look like a short. If this is from a text book, you can pretty safely assume that the capacitor's impedance is not zero when given schematics such as this. However, if you really want to be sure, you need to get some values, what is C, and what is the frequency of your source.