Why is there no voltage drop along a feedline
It is important to have your definitions straight. The source voltage is measured after the source's output impedance, as shown in below image:







simulate this circuit &ndash; Schematic created using CircuitLab

You are right that V1 in the image is double amplitude with respect to "Source Voltage", but that voltage is not part of the equation.

So with an ideal cable and matched in- and output impedances, the source voltage is equal to the voltage at the load end and with that reflection coefficient is 0:

Reflection coefficient

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{50 - 50 }{50 +50} = 0\$

VSWR

VSWR for a properly terminated line is defined as:

\$VSWR = \dfrac{V_{MAX}}{V_{MIN}} = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+0}{1-0} = 1\$

VSWR for an shorted line

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{0 - 50}{0 + 50} = -1\$

\$VSWR = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+1}{1-1} = \dfrac{2}{0} = \infty\$

Which basically means the voltage along the line varies between 2 and 0 times the source voltage. 

Voltage standing wave ratio is the ratio between the largest amplitude and the lowest amplitude found along the line. If the lowest amplitude nears zero, then VSWR will go to infinity.

VSWR for an open line

\$\varGamma = \dfrac{Z_L - Z_S}{Z_L + Z_S} = \dfrac{\infty - 50}{\infty + 50} = 1\$

\$VSWR = \dfrac{1+|\varGamma|}{1+|\varGamma|} = \dfrac{1+1}{1-1} = \dfrac{2}{0} = \infty\$

Which again basically means the voltage along the line varies between 2 and 0 times the source voltage.