Your assumptions for $\Phi$ are slightly weaker than what's needed for the contraction mapping principle, but the assumption that K is compact is stronger than is needed for the contraction mapping principle, so the idea is to repeat the proof of the contraction mapping principle using the compactness of K to fill in for the missing properties of $\Phi$.
First note that $\Phi$ is continuous (for any $\epsilon$ pick $\delta = \epsilon$, etc.).
Suppose $\Phi$ has no fixed point. Then $\forall (x) \; d(x, \Phi (x)) > 0$. Now since $d(\cdot , \Phi (\cdot))$ and $d(\Phi (\cdot), \Phi^2 (\cdot))$ are continuous (as compositions of continuous functions) and both are always non-zero, their ratio $r(\cdot):= d(\cdot, \Phi (\cdot))/d(\Phi (\cdot), \Phi^2 (\cdot))$ is also continuous. Since K is compact, there is an $x \in K$ where $r(x)$ is maximal. Let this maximal value be denoted by q. By hypotheses about $\Phi$, we have $q<1$.
Now for any point $x \in K$, consider the sequence $\left(\Phi^k (x)\right)_{k=0}^{\infty}$. We have $d(\Phi (x),\Phi^2 (x))<q d(x,\Phi (x))$. Inductively, $d(\Phi (x)^k,\Phi^{k+1} (x))<q^k d(x,\Phi (x))$. This is now the situation faced in the proof of the contraction mapping principle. The sequence has a cluster point y, which must be a limit point because for any $\epsilon$ we can find a large value k where both $d(\Phi^k (x),y)<\epsilon$ and $\forall (j>k)$:
$$
d(\Phi (x)^k,\Phi^j (x)) \\
\leq d(\Phi (x)^k,\Phi^{k+1} (x))+d(\Phi (x)^{k+1},\Phi^{k+2} (x))+...+d(\Phi (x)^{j-1},\Phi^j (x)) \\
\leq q^k + q^{k+1}+...+q^{j-1} \leq \frac{q^k-q^j}{1-q}\leq \epsilon
$$
y must be a fixed point, because $d(y, \Phi (y))\leq d(y,\Phi^k (x)) + d(\Phi (y),\Phi^k (x))\leq d(y,\Phi^k (x)) + q d(y,\Phi^{k-1} (x))$, which can be made arbitrarily small. This clearly contradicts the assumption that $\Phi$ had no fixed points.
Now showing that the fixed point is unique is easy: If there were two fixed points y and z, then $d(y,z)=d(\Phi (y), \Phi (z)) < d(y,z)$, contradiction.