I'm asked to decide if I should solve the system $$\dot{y}=\begin{pmatrix} -600 & 400 \\ 400 & -600 \end{pmatrix}y, \quad t\in[t_0,t_e], \quad y(t_0) = y_0$$ with either the explicit Euler method or the implicit Euler method.
Using the explicit Euler method I would get the updating scheme $$y_{n+1}=\begin{pmatrix} 1-600h & 400h \\ 400h & 1-600h \end{pmatrix}y_n$$ where the eigenvalues of the driving matrix is $$\lambda_1 = 401-600h,$$ $$\lambda_2 = -399 - 600h.$$ For the solution to be stable these need to be less than one which gives the conditions $$h\geq\frac{4}{6}=\frac{2}{3},$$ $$h\geq -\frac{4}{6}=-\frac{2}{3}.$$ The last condition doesn't say anything but the first condition seems restrictive since I can't choose $h$ as small as possible.
If I instead were to use the implicit Euler method I would get the updating scheme $$y_{n+1} = \begin{pmatrix} 1-600h & 400h \\ 400h & 1-600h \end{pmatrix}^{-1}y_n.$$ Now I can't solve for the eigenvalues of this system but I've heard the implicit Euler is unconditionally stable so it shouldn't matter.
So is the answer that I should choose implicit Euler because it is unconditionally stable or am I missing something? The order of consistency of both is $1$ so that should not matter.
Your computation of eigenvalues is wrong. Of the system matrix in $y'=Ay$ there is one eigenvalue $-200$ with eigenvector $\pmatrix{1\\1}$ and one eigenvalue $-1000$ with eigenvector $\pmatrix{1\\-1}$. These translate into eigenvalues $1-200h$ and $1-1000h$ of the "driving matrix" for the Euler method, requiring $h<0.002$ for stability.
For the implicit method there are no step size restrictions to get stability in the method, to get into the range where the error behaves like order 1 one still will need $500h\ll 1$.