The first part i) I can solve correctly, but I need some advice and intuition on how to solve the second part ii).
Here is the mark-scheme for the question: 
But for part ii) I do not understand their logic as shown in red i need to know why this must be the case for closest approach.
Could someone please talk me through this step by step in simple English as I have no idea the mark-scheme means?
Thank you,
with kind regards.
On
At any time, you can decompose your velocity vector into two components, one toward/away from $S$, and one perpendicular to the direction toward $S$. If the former component is nonzero, then you are not at closest approach: If the component is positive, your distance from the object is decreasing; if it is negative, your distance from the object is increasing. Either way, you're not at closest approach, which therefore requires that your distance toward the object is instantaneously zero.
From a physics perspective -- (here we go again, answering physics questions on a math site! ;) ) -- it may make more sense if you go into the inertial reference frame where the ship being approached is stationary.
This involves subtracting the (vector) velocity of the ship being approached from the velocities of both ships. Or put another way: pretend you're in a blimp that's directly above that ship at all times. You look down, you see that ship, and it's always in the same position.
Now that one ship is stationary, its velocity is zero, so it stays at the same place all the time. It's at a point. (The red point.)
The other ship moves along some straight line path (the blue path) past that point. (If the ship was moving at constant velocity in one inertial reference frame, it's moving at constant velocity in any other inertial reference frame.)
The minimum distance from the point to the line is a perpendicular. (The dashed line is my poor attempt at dropping the perpendicular from the point to the line.)