Geometrically, we want to find the values of the parameters $(\mu, \nu)$ such that the ellipse $ \frac{(x - (1-2\mu))^2}{(2\mu)^2} + \frac{y^2}{\nu^2} = 1 \tag{E} $ lies completely inside the unit circle $ x^2 + y^2 = 1. \tag{C} $ First of all, note that the left-most point of the ellipse -- namely, $(1-4\mu, 0)$ -- will lie inside the circle C if and only if $2 \mu \leqslant 1$. So we are left with finding the range of $\nu$ for a given $\mu$.
The ellipse will lie inside the unit circle if and only if, for all $x$ in the range $[1- 4\mu, 1]$, the height of the ellipse corresponding to the absicissa $x$ is smaller than or equal to the height of the circle: $ \nu^2 \left(1 - \frac{(x - (1-2\mu))^2}{(2\mu)^2} \right) \leqslant 1 - x^2. \tag{$\ast$}$
Here's a picture illustrating the final condition:

The picture shows the unit circle (in black), and three ellipses centered at $(1-2\mu, 0)$. The green ellipse is completely inside the circle, the red one exceeds it, and purple one is just inside the circle (touching it at $(1,0)$).
We pick an arbitrary “test point” $x$ is an arbitrary in the range $[1-4\mu, 1]$. The height of the ellipse at that point is denoted $y_e$ and that of the circle is denoted $y_c$. The inequality $(\ast)$ simply says that $y_e \leqslant y_c$ for all $x$ in the given range.
Our condition $(\ast)$ can be simplified as follows: $ \begin{array}{crl} \iff& \nu^2 \left( 1 - \left( 1 - \frac{1-x}{2\mu} \right)^2 \right) &\leqslant 1 - x^2 \\ \iff & \nu^2 \left( \frac{1-x}{\mu} - \frac{(1-x)^2}{4 \mu^2} \right) &\leqslant 1 - x^2 \end{array} $ Canceling a $1-x \geqslant 0$ factor, we have the equivalent condition $ \frac{\nu^2}{\mu} - \frac{\nu^2(1-x)}{4 \mu^2} \leqslant 1 + x \tag{$\dagger$} $ for all $x \in [1-4 \mu, 1]$.
Now in one direction, assuming $(\dagger)$ is satisfied, and plugging in $x = 1$ in it, we conclude that $\nu^2 \leqslant 2 \mu$.
In the other direction, we should prove that if $\nu^2 \leqslant 2 \mu \leqslant 1$, then $(\dagger)$ is satisfied for all $x \in [1-4\mu, 1]$. But this is straightforward: $ \begin{align*} \frac{\nu^2}{\mu} - \frac{\nu^2(1-x)}{4 \mu^2} &\leqslant 2 - \frac{(1-x)}{2 \mu} \\ &\leqslant 2 - \frac{(1-x)}{1} \\ &= 1+x. \end{align*} $
Thus we have showed that the ellipse lies inside the circle if and only if $\nu^2 \leqslant 2 \mu \leqslant 1$.