Let $V$ be a vector space over $\Bbb R$. As elaborated in the link in the comment, let $V_{\Bbb C} = V \oplus V$ denote the complexification of $V$, in which $v + iw = v \oplus w = (v,w)$. We define the conjugation map by
$$
J(v + iw) = v - iw
$$
For any $v,w \in V$. Note that $J$ is $\Bbb R$-linear and that for any $\lambda = a+bi$, $x = v + iw$, we have:
$$
J(\lambda x) =
J[\lambda(v+iw)] = \overline{\lambda}(v - iw) = \overline{\lambda}J(v + iw) = \overline{\lambda}J(x)
$$
which I will let you verify. In other words, $J$ is antilinear.
Now, if $T:V \to V$ is linear, then the unique $\Bbb C$-linear extension to $V_{\Bbb C}$ is given by
$$
\tilde T(v + iw) = T(v) + iT(w)
$$
It follows that
$$
\tilde TJ(v + iw) = T(v - iw) = T(v) - iT(w) = J\tilde T(v + iw)
$$
That is, if $\tilde T:V_{\Bbb C} \to V_{\Bbb C}$ is the $\Bbb C$-linear extension of an $\Bbb R$-linear map, then $\tilde TJ = J\tilde T$. With that, we may proceed:
Theorem: Suppose that $\tilde T:V_{\Bbb C} \to V_{\Bbb C}$ is the $\Bbb C$-linear extension of an $\Bbb R$-linear map on $V$. If $\lambda$ is an eigenvalue of $\tilde T$ with eigenvector $v \in \Bbb V_{\Bbb C}$, then $\overline{\lambda}$ is an eigenvalue of $\tilde T$ with eigenvector $\overline{v} = Jv$.
Proof: Note that
$$
\tilde T\overline{v} = \tilde T(Jv) = J(\tilde Tv) = J(\lambda v) = \overline{\lambda}J(v) = \overline{\lambda} \overline{v}
$$
as desired.
Or, to more closely mirror your referenced proof:
$$
(T - \lambda I)v = 0 \implies\\
J(T - \lambda I)v = 0 \implies\\
(JT - J(\lambda I))v = 0\implies\\
(TJ - \overline{\lambda}IJ)v = 0 \implies\\
(T - \overline{\lambda} I)(Jv) = 0 \implies\\
(T - \overline{\lambda} I)\overline v = 0
$$
There is essentially no condition on the space on which your functions live that will ensure existence of any eigenfunctions. For example, for $X=[a,b]$ and $V$ the space of real-valued (or complex-valued, hardly matters...) functions on $X$, and $L$ the multiplication-by-$x$ operator... there are no eigenfunctions at all (unless $a=b$) [EDIT: among continuous or $L^2$ or $L^p$ functions]. That is, unless the physical space is a finite set (or has a weird-enough topology so that there aren't very many continuous functions), some very simple, non-pathological operators fail to have any eigenvalues at all (EDIT: among continuous or $L^2$ or...].
Another non-pathological example that shows that it is not generally reasonable to expect eigenvalues is the Laplacian on the real line. Fourier inversion shows that everything in $L^2$ is a superposition of generalized eigenvalues [EDIT: oop, eigenfunctions] (the exponentials), but not a sum, and those eigenvalues are not in the space itself. EDIT: Fourier inversion expresses (e.g.) a Schwartz function $f$ as
$$
f(x) \;=\; \int_{\mathbb R} e^{2\pi i\xi x}\;\widehat{f}(\xi)\; d\xi
$$
where $\widehat{f}$ is the Fourier transform of $f$ Thus, $f$ is a superposition (=integral) of eigenfunctions (the functions $x\to e^{2\pi i\xi x}$ for $\Delta=\partial^2/\partial x^2$.
Yes, in the context of Sturm-Liouville problems (see also Fredholm alternative), the point is that the inverse of the differential operator (with boundary conditions) is a compact self-adjoint operator on a Hilbert space of functions, and the eigenvalues are in bijection by $\lambda \leftrightarrow \lambda^{-1}$, etc. The only general class of operators on infinite-dimensional spaces with a clear, simple, and happy spectral theory are compact self-adjoint, or unbounded operators which have compact self-adjoint operators as inverses/resolvents...
Thus, for example, the Laplace-Beltrami operator on a compact Riemannian manifold does provably have compact resolvent, so $L^2$ has an orthonormal basis of eigenfunctions.
EDIT: I'd also add that one almost surely wants a reasonable topology on the space of functions, related to the topology on the underlying physical space. And we'd want some sort of completeness on the space of functions, else we'd potentially lose eigenfunction/values for silly reasons.
Best Answer
Let $\mathcal{H}$ be an infinite-dimensional Hilbert space and let $S$ be any operator on $\mathcal{H}$ that has no eigenvalues [for example, take $\mathcal{H} = L^2[0, 1]$ and let $S$ be the operator on $\mathcal{H}$ defined by $(Sf)(x) = xf(x)$.] Now define an operator $T$ on $\mathcal{H} \times \mathcal{H}$ by $$T(f, g) = (0, Sg).$$ Then $0$ is the only eigenvalue of $T$, but $T$ is not nilpotent.