Suppose $f:\mathbb{R}\to\mathbb{R}$ is a smooth, positive, bounded function on $\mathbb{R}$. Construct a real-valued continuous function $u$ on $\overline{\mathbb{H}}$ which are harmonic on $\mathbb{H}$, such that $u|_{\mathbb{R}} = f$ and $u(i) = 0$, where $\mathbb{H}$ is $\{z : Im(z) > 0\}$ }
I don't know how to construct, and I also feel confused: Suppose $u$ exists, then we consider the function $u\circ h$, where $h=\frac{i(w+1)}{1-w}$. So $u\circ h$ is defined on $D$ with positive value on $\partial D$, and $u\circ h(0)=u(i)=0$, then by mean value property, we deduce a contradiction.
This is a Schwarz problem.
Just as you mentioned, first of all, find a holomorphic function $g$ in $\mathbb{H}$, such that $\Re\left(g\right)$, when restricted on $\mathbb{R}$, is identical to $f$. This can be done by $$ g(z)=\frac{1}{\pi i}\int_{\mathbb{R}}\frac{f(x)}{x-z}{\rm d}x. $$ If $f$ is merely bounded but not integrable, you may use $$ g(z)=\frac{1}{\pi i}\int_{\mathbb{R}}\frac{f(x)\left(z-z_0\right)}{\left(x-z\right)\left(x-z_0\right)}{\rm d}x $$ instead, where $z_0\in\mathbb{H}$ could be arbitrarily chosen.
So far, this $g$ yields that $\Re\left(g\right)$ is bounded in $\mathbb{H}$, which satisfies the mean-value property.
Secondly, consider $$ h(z)=g(z)+iKz, $$ where $K\in\mathbb{R}$ will be determined by the constraint $\Re\left(h\right)(i)=0$. This $\Re\left(h\right)$ will be the very function you are looking for.