By my read of the question, I don't see why this is even correct. It does depend on how one interprets that "H is not computation resistant for every key K". I assume that this just means that $H$ is not computation resistant for a randomly chosen key $K$. Otherwise, I'm not sure what it would mean.
According to this understanding, I don't see why the statement is true. From here on, I will use the term "a secure MAC" rather than "computation resistant", since this is more standard terminology.
Now, the straightforward way you would try to prove this is to find a collision in $H_K$ and use this to break the MAC. However, the fact that $H_K$ is not collision resistant for a fixed $K$ means that when you are given the function description, you can find a collision. However, this doesn't mean that you can also do it when you are not given the function description. To be very concrete, define $H_K$ as follows. Parse $K$ into a key for a PRF and into the description of a group where the discrete log is assumed to be hard, along with a generator $g$ and random value $a$ and value $h=g^a$. The function works by first hashing the message using $hash(x_1\|x_2)=g^{x_1}\cdot h^{x_2}$ (using Merkle-Damgård based on this compression function, you get a full-blown hash function) and then applying the PRF. Now, if you know the value $a$, then it's easy to find collisions. Thus, $H_K$ is not collision resistant for any fixed key, since in the fixed-key setting you know $K$. However, in the standard MAC setting where you don't know $K$ and so don't know $a$, this will be a secure MAC (under the assumption that the discrete log problem is hard in this group).
So, I must be misreading the question somewhere. Either there is something in the quantifiers with the "for every key $K$" that I am misreading, or the intention of what a fixed-keyed hash function is is different to what I understand. Or there's an error in the question...
In any case, I wrote this out in detail since I think it has value.
