I'm hoping someone could verify my answer to the following problem:
Consider a function $f$ that is continuous for $Im(z) \geq 0$ and analytic for $Im(z) > 0$. Furthermore, assume that $f$ send the real axis to purely imaginary numbers. Construct an analytic continuation for $f$ into $\mathbb{C}$.
Well, using the reflection principle, we have $\overline{f(z)} = f(\overline{z})$ or $f(z) = \overline{f(\overline{z})}$. Thus, the continuation $F$ can be defined by
$$ F(z) = \begin{cases} f(z), & Im(z) > 0 \\ \overline{f(\overline{z})}, & Im(z) \leq 0 \end{cases} $$
Yes. This is true by the Schwarz reflection principle, to which you alluded.
The Schwarz reflection principle states that if $f$ is analytic on an open region in the upper-half plane and purely real on the real axis, then $\bar{f(\bar{z})}$ is analytic in the reflection.
Suppose you wanted to show this.
Note that $f(z) = \bar{f(\bar{z})}$ on the real axis.
Let $f(z) = u(x,y) + iv(x,y)$. Let $\bar{f(\bar{z})} = u(x, -y) - iv(x, -y) = U(x,y) + i V(x,y)$. Then $\frac{\partial U}{\partial x} = \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = \frac{\partial V}{\partial y}$ and $\frac{\partial U}{\partial y} = -\frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} = -\frac{\partial V}{\partial x}$. Thus $\bar{f(\bar{z})}$ satisfies Cauchy's equations.
When asking if the function $F(z) = \begin{cases} f(z) & \text{Im} z \ge 0 \\ \bar{f(\bar{z})} & \text{Im} z < 0\end{cases}$ is actually an analytic continuation of $f(z)$ into $\mathbb{C}$, you might be concerned about analyticity on the real line, where $f$ is only specified to be continuous.
You can use Morera's theorem to show that $F(z)$ is entire. Take contours $C_1, C_2$ in the upper and lower half plane respectively. In the limit as the contours approach the boundary at the real axis (where $f(z) = \bar{f(\bar{z})}$ is continuous), the two contours "cancel out" along the real axis--we can then form a contour integral around the union of the two regions above an below the real axis which integrates to zero, thus showing $F(z)$ is analytic in the union by Morera's theorem.