I would start by seeing roughly how close the lines are to that point. Probably most of them could be eliminated by comparing them that way. Then, if more than one are left, get more technical and exact, if necessary.
(A) Closest integer points are (0,0) and (1,1). Those two are each 1 away from (1,0). This line has a slope of 1 and rises at a 45 degree angle. Between those two integer points is a spot on the line that is diagonally closer than 1 away from (1,0).
So < 1.
(B) The closest point on this line is (1, 1). It's 1 away from (1, 0), and as soon as we move to the left or right of it we go further from (1, 0).
So = 1, and A is closest so far.
(C) This is the same as y = -x + 3.
When y = 0, x = 3. (3, 0) is 2 away from (1,0).
When x = 1, y = 2. That's also 2 away from (1,0).
There is a closer point between those two points, but since y = -x + 3 has a slope of -1, it is a 45 degree line, like y = x, and y = x is only one away at integer points. So A is still closer.
(D) x = 2 is 1 away at its closest point, (2,0), similarly to how y = 1 is.
A is still winning.
(E) y = -x - 1.
When x = 1, y = -2, 2 away from (1,0).
When y = 0, x = -1, 2 away from (1,0).
So, like the line in C, this has a 45 degree or - 1 slope, and passes further way than y = x.
So there was no need to get very technical or calculate the exact distances.
A wins.