To do this, fix any (u,v) in the disc with center (x,y) and radius epsilon. Then
(1) (u - x)^2 + (u - y)^2 < epsilon^2.
Since (u - y)^2 is nonnegative, it follows that (u - x)^2 <= (u - x)^2 + (u - y)^2, and hence from (1) that (u - x)^2 < epsilon^2. Taking the square root of both sides and using the fact that epsilon > 0, we deduce that |u - x| < epsilon.
The inequality |u - x| < epsilon implies that (2) u - x < epsilon and that (3) -epsilon < u - x. From (2) and the fact that epsilon < 1 - x by definition, we deduce that u - x < 1 - x and hence u < 1. From (3) and the fact that epsilon < x by definition (which implies that -x < -epsilon) we deduce that -x < u - x, which implies that u > 0. So 0 < u < 1.
Similarly, since (u - x)^2 is nonnegative, (1) implies that (u - y)^2 < epsilon^2 and hence that |u - y| < epsilon. If we repeat the previous paragraph with v's and y's in place of u's and x's we deduce that 0 < v < 1.
Since 0 < u < 1 and 0 < v < 1 we deduce that (u,v) is in G and hence, since (u,v) was arbitrary, that the disc with center (x,y) and radius epsilon is a subset of G. Since (x,y) was an arbitrary point in G, this implies that G is open.
It is worth pointing out that in "real life" (real mathematical life, anyway) one generally does not prove that sets like G are open right from the definition. One usually has a large library of theorems about open sets, and uses them instead. For example,
(a) G is the intersection of the four sets {(x,y) in R^2: x < 1}, {(x,y) in R^2: x > 0}, {(x,y) in R^2: y < 1}, and {(x,y) in R^2: y > 0}
(b) The inverse image of an open set under a continuous function is open.
(c) The functions f and g from R^2 to R given by f(x,y) = x and g(x,y) = y are continuous.
(d) {(x,y) in R^2: x < 1} and {(x,y) in R^2: x > 0} are the inverse images of the open sets (-oo, 1) and (0,oo) under f, and {(x,y) in R^2: y < 1} and {(x,y) in R^2: y > 0} are the inverse images of the open sets (-oo,1) and (0,oo) under g. And hence by (c) and (b), each of the four sets listed in (a) are open.
(e) A finite intersection of open sets is open.
(f) So G is open.
Each of the steps (a), (b), (c), (d), (e) is actually very easy to verify.