Answer
procedure insert ($x:$ integer to be inserted, $a_1, a_2, ..., a_n:$ list of integers arranged in increasing order)
for $i:=1$ to $n$
$\space\space\space$ if $x
Work Step by Step
The algorithm attempts to locate the position at which to insert the element $x$ in the arranged set. Two main points have to be made:
First, notice that when iterating though the loop, we check if the current element is strictly greater than $x.$ As soon as that is true, we insert $x$ before that element. This is because we know that up to that point, not a single element we met was strictly greater than $x$ and hence all were either less than or equal. Hence, placing $x$ will retain the order of the integers.
The second line is necessary for the edge case when $x$ is greater than all the elements of the set. If you trace the loop, you would notice that $x$ is never inserted into the set when it is greater than every element. This is why we have to explicitly add the final line.
