Is this is a misprint? I was given a question which said that $H\leq K \leq G$ meant that $C_G(K)\leq C_G(H) $.
This doesnt seem right though, my gut is telling me it should say that K is a subgroup of H instead.( ofcourse the gut can never be trusted in such things).
So is this a misprint ? ( I want to work it out myself I just don't want to spend hours on a proof that may have been posed incorrectly ) 
 A: Smaller group means weaker conditoin:
$$x\in C_G(K)\implies \forall y\in K\colon xy=yx\implies \forall y\in H\colon xy=yx\implies x\in C_G(H).$$
A: No, the statement is correct. $H$ is a subgroup of $K$. So if an element commutes with all elements even in $K$ then for sure it commutes the all the elements of $H$. 
A: In addition to the answers by Mark and Hagen von Eitzen, I just want to make a couple of remarks, since something funny is going on here with the centralizer of the centralizer of the centralizer .... There are two things one should remember and can be proved hands-down or by tautology ((b) below!). Let $G$ be a group and $H$ and $K$ subgroups. 
(a) If $H \subseteq K$, then $C_G(K) \subseteq C_G(H)$. 
(b) $H \subseteq C_G(C_G(H)).$
Now if you plug in $C_G(H)$ in stead of $H$ in (b) you get $C_G(H) \subseteq C_G(C_G(C_G(H)))$. On the other hand, (a) applied to (b) yields $C_G(C_G(C_G(H))) \subseteq C_G(H)$. So we get: $C_G(H)=C_G(C_G(C_G(H)))$. 
Now put $C_{G,0}(H)=H$ and $C_{G,i}(H)=C_G(C_{G,i-1}(H))$ for $i \geq 1$. Then it follows that for all $j \geq 1$ $C_G(H)=C_{G,2j-1}(H)$ and $C_G(C_G(H))=C_{G,2j}(H)$. So these "centralizer series" stabilize very quickly on the odd and even indices.
Note that $H$ does not have to be equal to $C_G(C_G(H))$: take a group $G$ with a non-cyclic center, and let $g \in Z(G)$ be a non-trivial element. Put $H=\langle g \rangle$, then $H \subsetneq C_G(C_G(H))=Z(G)$.
