Answer: D

Solution:

Let the integers in increasing order be $a,b,c,d,e.$ The median is $c,$ and the unique mode is $c+2.$

Because the mode is larger than the median and is unique, the last two entries must both be $c+2,$ so the list is $a,b,c,c+2,c+2.$

The mean is $c-2,$ so $$\begin{aligned} &\frac{a+b+c+(c+2)+(c+2)}{5} \\ &\quad = c-2. \end{aligned}$$ Hence $a+b+3c+4=5c-10,$ so $a+b=2c-14.$

To keep the mode unique, $a$ and $b$ must be distinct positive integers, both less than $c.$ Since $a+b=2c-14$ is even, the smallest such sum is $1+3=4,$ so $2c-14\ge4,$ giving $c\ge9.$

The smallest possible mode is therefore $c+2=11,$ and it is attainable with $1,3,9,11,11.$

Thus, the answer is D .