First of all, the reflection $(x', y')$ of $(x_0, y_0)$ in $L$ is given by
$$x' = \frac{(1 - m^2)x_0 + 2my_0}{1 + m^2}, \quad y' = \frac{2mx_0 + (m^2 - 1)y_0}{1 + m^2}.$$
This is so because
$$x' + x_0 = \frac{2(x_0 + my_0)}{1 + m^2}, \quad y' + y_0 = \frac{2m(x_0 + my_0)}{1 + m^2}.$$
Thus the reflection of $(1, 1)$ in $L$ has coordinates
$$x' = \frac{1 + 2m - m^2}{1 + m^2}, \quad y' = \frac{m^2 + 2m - 1}{1 + m^2}.$$
Next, if $m+1 \neq 0$, the slope of the line joining $(x', y')$ to $(-1, 1)$ is given by
$$\mu = \frac{y' - 1}{x' + 1} = \frac{m^2 + 2m - 1 - (1 + m^2)}{1 + 2m - m^2 + 1} = \frac{m - 1}{m + 1}.$$
Hence, the line joining $(-1, 1)$ and $(x', y')$ has equation
$$(m+1)(y-1) = (m-1)(x+1).$$
This line meets $L$ at the point $P$ with coordinates
$$x = \frac{2m}{m^2+1}, \quad y = \frac{2m^2}{m^2+1}.$$
Since
$$x^2 + (y - 1)^2 = \frac{4m^2}{(m^2 + 1)^2} + \left(\frac{m^2 - 1}{m^2 + 1}\right)^2 = \frac{4m^2 + m^4 - 2m^2 + 1}{(m^2 + 1)^2} = 1,$$
the point $P$ traverses an arc of the circle of unit radius centred at $(0, 1)$.

Second Solution:
To solve this problem using synthetic geometry, we denote the unit circle centred at $(0, 1)$ by $C$ and let $A = (0, 0)$, $B = (1, 1)$ and $C = (-1, 1)$. The line $L$ intersects $C$ at $A$ and a second point $P$. Let $B'$ be the intersection point of the line $PC$ with the line which is perpendicular to $L$ and passes through $B$. Let $M$ be the intersection point of $PA$ and $BB'$.

The statement we have to show is then equivalent to $|BM| = |MB'|$. We obtain this as follows. Because the two arcs $CA$ and $AB$ are equal (both are quarter circles or $\pi/2$), we have $\angle CPA = \angle APB$ if $P$ is on the arc $BC$ not containing $A$ and we have $\angle CPA = 180^\circ - \angle BPA = \angle MPB$ if $P$ is on the arc $AB$ not containing $C$. The case where $P$ is on the arc $AC$ not containing $B$ is similar and if $P$ coincides with $A$, $B$ or $C$ the statement is obvious. Because $\angle PMB' = \angle BMP$ (both are right angles), the two triangles $BPM$ and $PB'M$ share the common side $PM$ and have equal adjacent angles, hence are congruent. This shows $|BM| = |MB'|$.