This isn't an answer, but evidently I don't have enough points to leave a "comment." I don't have a reference on Sturm-Liouville problems handy. However option (2) doesn't make sense. If there is one eigenfunction, there are infinitely many for the same eigenvalue: just take your eigenfunction $u$ and multiply by any non-zero real $\alpha$. Also, (3) implies (4): if there are two linearly independent eigenfunctions, you can produce two orthogonal functions (spanning the same subspace) using a Gram-Schmidt type argument.
Hope this clarifies a couple things.
Related problems: (I), (II). First solve the differential equation
$$ X(x)=c_1\,\sin \left( \sqrt {\lambda}x \right) + c_2\,\cos\left( \sqrt {\lambda}x \right) .$$
Applying the boundary conditions to the solution results in the two equations
$$ { c_1}\,\sin \left( \sqrt {\lambda}\pi \right) +{c_2}\,\cos\left( \sqrt {\lambda}\pi \right)
=0 \rightarrow (1) $$
$$ {c_1}\,\sqrt{\lambda} = 0 \rightarrow (2), $$
where $c_1$ and $c_2$ are arbitrary constants. From (2), we assume $\lambda\neq 0$, then we will have $c_1=0.$ Substituting $c_1=0$ in (1) gives
$$ {c_2}\,\cos\left( \sqrt {\lambda}\pi \right)
= 0 \implies \cos\left( \sqrt {\lambda}\pi \right)=0 \implies \sqrt{\lambda} = \frac{2n+1}{2} $$
$$ \implies \lambda = \frac{(2n+1)^2}{4},\quad n=0,1,2,3\dots $$
I will leave it here for you to finish the task. Note that, $\lambda = 0 $ is a special case. Subs $\lambda=0$ in the diff. eq. and follow the above technique and see what you get.
Best Answer
The differential equation you seek is
$$y''+\lambda y = 0$$
which had the general solution
$$y(x) = A \cos{\sqrt{\lambda} x} + B \sin{\sqrt{\lambda} x}$$
The boundary value given by $y'(0)=y(0)$ implies that $A = B \sqrt{\lambda}$. The boundary value given by $y'(1) + y(1) = 0$ implies that
$$-A \sqrt{\lambda} \sin{\sqrt{\lambda}} + B \sqrt{\lambda} \cos{\sqrt{\lambda}} + A \cos{\sqrt{\lambda}} + B \sin{\sqrt{\lambda}}=0$$
Using the relation derived from the first boundary condition, we get
$$-\lambda \sin{\sqrt{\lambda}} + \sqrt{\lambda} \cos{\sqrt{\lambda}} + \sqrt{\lambda} \cos{\sqrt{\lambda}} + \sin{\sqrt{\lambda}}=0$$
or, after a little algebra,
$$\tan{\sqrt{\lambda}} = \frac{2 \sqrt{\lambda}}{\lambda-1}$$
Here is a plot of the LHS and RHS so you can see the intersections and hence, the square roots of the eigenvalues. Clearly, for the first few, they must be computed numerically. For the $k$th eigenvalue, where $k$ is large,
$$\lambda_k \sim k^2 \pi^2$$
The eigenfunction (unnormalized) corresponding to that eigenvalue is
$$y_k(x) = \sqrt{\lambda_k} \cos{\sqrt{\lambda_k} x} + \sin{\sqrt{\lambda_k} x}$$
The Green's function may then be found from the eigenvalues/eigenfunctions as follows:
$$G(x,x') = \sum_{k=1}^{\infty} \frac{y_k(x) y_k(x')}{\lambda-\lambda_k}$$
where $G$ satisfies
$$\frac{d^2}{dx^2} G(x,x') + \lambda G(x,x') = \delta(x-x')$$