This answer does not use complex analysis per se, but a basic property of the Lebesgue integral, the dominated convergence theorem. Sorry! I thought you might find it useful nonetheless.
To show that $u$ is harmonic, choose a point $z_0\in U$ and observe that each $u_n$ satisfies the mean-value property at $z_0$:
$$
u_n(z_0) = \frac{1}{2\pi r}\int_{|z-z_0|=r} u_n(z), \text{ for all sufficiently small $r>0$.}
$$
Since each $u_n$ is dominated in modulus by $1$, the dominated convergence theorem allows us to let $n$ approach infinity and interchange limit with integral. Then
$$
u(z_0) = \frac{1}{2\pi r}\int_{|z-z_0|=r} u(z), \text{ for all sufficiently small $r>0$.}
$$
Consider $x_0 \in \partial \mathbb{D}$ and an open disk $B_r(x_0)$ of radius r where $|x -x_0| <r$.
We have
$$\tag{*}|u(x) - u(x_0)| \\ \leqslant \left|\int_{\partial \mathbb{D} - B_r(x_0)}f(y) [\ln |x-y|-\ln|x_0-y|] \, dS(y) \right| + \left|\int_{\partial \mathbb{D} \cap B_r(x_0)}f(y) \ln |x-y| \, dS(y) \right|+ \left|\int_{\partial \mathbb{D} \cap B_r(x_0)}f(y) \ln |x_0-y| \, dS(y) \right|$$
Choosing $|x - x_0|$ (and $r$) sufficiently small we can make each term on the RHS of (*) smaller than $\epsilon/3$.
For the second and third terms we can use the same argument that proves the integral exists.
For the first term note that $||x-y| - |x_0-y|| \leqslant |x - x_0|$. Since $x,x_0 \neq y$ for $y \in \partial \mathbb{D} - B_r(x_0)$ and the logarithm function is continuous, we have for sufficiently small $|x-x_0|$ and $M = \sup_{y \in \partial \mathbb{D}} f(y),$
$$| \ln |x-y| - \ln |x_0-y|| < \frac{\epsilon}{6\pi M}, $$
which implies that the first term on the RHS of (*) is less than $\epsilon/3$.
To be more precise, the argument rests on first choosing $r$ such that the contribution from the second and third terms on the RHS of (*) is smaller than $2\epsilon/3$. Then we would show that $|\ln|x -y| - \ln|x_0 - y|| \to 0$ as $x \to x_0$ uniformly for $y \in \partial \mathbb{D} - B_r(x_0)$
Best Answer
For any bounded measurable function $f\in L^\infty(S^1)$, the Poisson integral $P[f]$ gives a bounded harmonic functions in $B_1$. This function is continuous (equivalently, uniformly continuous) if and only if $f$ itself is continuous.
Here you can find what the Poisson kernel is and how it's used to build harmonic functions in the ball.
Said otherwise, $f\mapsto P[f]$ is a linear isometry of $L^p(S^1)$ onto $h^p(B_1)$. We used the case $p=\infty$, where $h^\infty(B_1)$ stands for the the bounded harmonic functions in the ball. You can read more about it here.