Let $K$ be compact with $K \subset \mathbb D$. Also let $a \in K$ and $u$ be a positive harmonic function on $\mathbb D$. 
Let $K$ be compact with $K \subset \mathbb D$. Also let $a \in K$ and $u$ be a positive harmonic function on $\mathbb D$. Prove there exists a constant $C = C(a, K) > 0$ such that for all $z \in K$ 
  $$
\frac{u(a)}{C} \le u(z) \le Cu(a).
$$
  Edit: The constant $C$ should depend on $a$ and $K$ only, not $u$. 

I think Harnack's inequality should be used. 
Since $K$ is compact inside an open set there exists $\delta = d(K, \partial \mathbb D) > 0$. Then we can take a disc centered at $a$ of radius $1 - \delta < R < 1$. So Harnack's inequality can be applied on $\overline{D(a, R)}$ however we don't know if this disc covers all of $K$. I'm stuck here. 
I've noted that we can take $K \setminus D(a, R)$ which is a compact set, thus we can take the open cover
$$
O = \{ B(z, R) : z \in K \setminus D(a, R) \}
$$
and obtain a finite subcover $O_N = \{ B(z_i, R) : 1 \le i \le N \}$. Not sure if this is helpful.
 A: You can prove this for compact subsets of any connected open set by an argument like what you started. For the disk you can cheat and give a simpler argument, using the fact that $u$ is the Poisson integral of  some positive measure on the circle. (Which of course is how Harnack's inequality is proved in the first place -- look back at that proof and you see it actually gives a proof of this.)
To be explicit: There exists a positive measure $\mu$ on $\Bbb T$ such that $$u(z)=\int_{\Bbb T}P_z(t)\,d\mu(t),$$where $P_z(t)$ is the Poisson kernel. Given $K$ compact, you can show that there exist $c>0$ and $C>0$ such that $$c\le P_z(t)\le C\quad(z\in K, t\in\Bbb T).$$Hint for that: there exists $r<1$ so $|z|\le r$ for every $z\in K$. This shows that $$c u(0)\le u(z)\le C u(0)\quad(z\in K),$$so that if $a\in K$ then $$u(z)\le  C u(0)\le \frac Cc u(a),\quad(z\in K)$$and similarly $u(z)\ge\frac cC u(a)$.
If you don't know that fact about positive harmonic functions and measures: for $0<r<1$ define $u_r(z)=u(rz)$. It's enough to prove the inequality with $u_r$ in place of $u$, since $u_r\to u$ pointwise as $r\to1$. But $u_r$ is continuous on the closed disk, so you can apply the Poisson fomrmula for $u_r$ as above.
A: Since $K$ is compact und $u$ is continuous on $K$, there are $c,d \in K$ such that
$u(c) \le u(z) \le u(d)$ for all $z \in K$.
For $n \in \mathbb N$ we have $\frac{u(a)}{n} \to 0$ and $nu(a) \to + \infty$ as $n \to \infty$.
Hence there is $N \in \mathbb N$ such that
$\frac{u(a)}{N} \le u(c)$ and $u(d) \le Nu(a)$.
Now take $C(a,K)=N$.
