If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity This is a reworking of a previous question here which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.
I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:

If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|<\rho$ when $u(0)$ is properly defined.

My Solution So Far: 
Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected. 
If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 < |z| < \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.
One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$
in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 < |z| < \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?
Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant...
 A: Here is proof based on the contents of the book until that point:
Theorem 20 in the text states that for $0<r<1$ $$\frac{1}{2 \pi} \oint_{|z|=r} u\, \mathrm{d} \theta=\alpha \log r+\beta $$
Where $\alpha=\oint_{|z|=r} r \frac{\partial u}{\partial r} \, \mathrm{d} \theta=\oint_{|z|=r}*\mathrm{d}u$. ($*\mathrm{d}u=-\frac{\partial u}{\partial y} \mathrm{d} x+ \frac{\partial u}{\partial x} \mathrm{d} y$ is the conjugate differential.)
Let us take $r \to 0^+$ in the theorem: since $u$ is bounded the integral on the left tends to zero, and therefore the $\log$ term must have coefficient zero. So that $\oint_\gamma *\mathrm{d} u$ is zero for every circle with radius $0<r<1$. $|z|=\frac{1}{2}$ is such a circle, and in fact it is a homology basis for the punctured disk. Thus the integral $\oint_\gamma * \mathrm{d}u$ vanishes for every cycle $\gamma$ in the punctured disk, and we can define a harmonic conjugate $v$ there.
The function $f=u+iv$ is analytic in the punctured disk, and it has bounded real part around $z=0$. According to exercise 5 on page 130 we find that the singularity at $0$ is removable, so that we can extend $f$ to an analytic function in the whole disk, and $u$ is extended harmonically that way.
