(1) Let's start by considering the standard equation of an ellipse centered at the origin and with its axes aligned with the coordinate axes which is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b > 0$.
Since the length of the minor axis is 2, we have $2b = 2$ which implies $b = 1$. Additionally, we know that the eccentricity $\varepsilon = \frac{c}{a} = \frac{2\sqrt{5}}{5}$, which leads us to $\frac{c^2}{a^2} = \frac{4}{5}$.
We know that in an ellipse $a^2 = b^2 + c^2$, so $a^2 = 1^2 + c^2 = 5$. Thus, the equation of ellipse C is $\boxed{\frac{x^2}{5} + y^2 = 1}$.

(2) To prove the second part, let's define the coordinates of the points A, B, and M as A$(x_1, y_1)$, B$(x_2, y_2)$, and M$(0, y_0)$, respectively. We also know that the right focus F has coordinates $(\sqrt{a^2 - b^2}, 0)$, so the coordinates of F are $(2, 0)$ since $a^2 = 5$ and $b^2 = 1$.
The slope of line l exists and let's denote it by $k$, so the equation of line l can be written as $y = k(x - 2)$.

Substituting the equation of line l into the equation of ellipse C and eliminating $y$ and simplifying gives us $(1 + 5k^2)x^2 - 20k^2 x + 20k^2 - 5 = 0$.
Consequently, our resulting system is $x_1 + x_2 = \frac{20k^2}{1 + 5k^2}$ and $x_1 x_2 = \frac{20k^2 - 5}{1 + 5k^2}$.

Since $\vec{MA} = \lambda_1 \vec{AF}$ and $\vec{MB} = \lambda_2 \vec{BF}$, we substitute the coordinates of the points A and B and obtain $\lambda_1 = \frac{x_1}{2 - x_1}$ and $\lambda_2 = \frac{x_2}{2 - x_2}$. Therefore,
\begin{align*}
\lambda_1 + \lambda_2 &= \frac{x_1}{2 - x_1} + \frac{x_2}{2 - x_2} \\
&= \frac{2(x_1 + x_2) - 2x_1 x_2}{4 - 2(x_1 + x_2) + x_1 x_2} \\
&= \frac{2\cdot \frac{20k^2}{1 + 5k^2} - 2\cdot \frac{20k^2 - 5}{1 + 5k^2}}{4 - 2\cdot \frac{20k^2}{1 + 5k^2} + \frac{20k^2 - 5}{1 + 5k^2}} \\
&= \frac{40k^2 - (40k^2 - 10)}{(4+5k^2) - (40k^2)} \\
&= \frac{10}{5k^2+4-40k^2} \\
&= \frac{10}{4 - 35k^2} \\
&= -10 \quad (\text{as } 35k^2 > 0)
\end{align*}

Thus, $\lambda_1 + \lambda_2$ is a constant, specifically $\boxed{-10}$.