why is g cintinous?

g(x+y) = g(x) g(y)

at x=0, g(0+y) = g(y) = g(0) g(y)

so g(0) = 1

If g is continuous at x=0, then g(0+y) = g(y) is continuous for all y.

Since any y can be decomposed into y=a+b, where a,b are real numbers, then g(y) = g(a) g(b) must be continuous, so g must be continuous everywhere.

If g(a) = 0 for some a in R, then write a = x+y where x,y are real numbers, then

g(a) = g(x+y) = g(x) g(y) = 0

then either g(x) = 0 or g(y) = 0 (or both).

that implies that g(x) = 0 for all x in R.