If I'm interpreting correctly, this is the troublesome sentence:

If one particular wavevector $k$ is contained in a $\psi$, then all other wavevectors in the Fourier expansion of this $\psi$ will have the form $k+G$, where $G$ is any reciprocal lattice vector.

Now, you ask why. The answer is that, well, Charles Kittel is lying here. They're not looking for just any solutions, they're looking for a helpful set of solutions. Let's take a step back and derive the central equation $$ (\lambda_k - \epsilon)C(k) + \sum_{G}U_GC(k-G) = 0. $$

Starting at the previous equation: $$ \sum_k \lambda_k C(k) e^{ikx} + \sum_G\sum_k U_GC(k)e^{i(k+G)x} = \epsilon\sum_kC(k)e^{ikx} $$ The second term on the right side can be rearranged: $$ \sum_G\sum_k U_GC(k)e^{i(k+G)x} = \sum_k\left(\sum_GU_gC(k-G)\right)e^{ikx} $$ and then we get to rearrange the previous equation to have a sum over $k$ on one side, zero on the other: $$ \sum_k \left((\lambda_k - \epsilon) C(k) + \sum_GU_gC(k-G)\right)e^{ikx} = 0 $$ Only way for this to be true is for the coefficients to each be zero: $$ (\lambda_k - \epsilon) C(k) + \sum_GU_gC(k-G) = 0 $$ Once we have a $k_0$ value with $C(k_0)\neq 0$, we have all of the $C(k_0-G)$ coefficients determined by the central equation. Moreover, saying that $C(k\neq k_0-G) = 0$ will yield a valid wavefunction with energy $\epsilon$.

But what if $C(k\neq k_0-G)\neq 0$? Well, the honest answer is that we end up with the same result: a wavefunction with energy $\epsilon$. There's nothing actually stopping us from having all values of $k$ represented.

But the helpful part of using $C(k\neq k_0-G) = 0$ is that it organizes the solutions into a set of $\psi_k$ with wonderful properties:

1. The solutions $\psi_k$ form a basis.
2. The solutions $\psi_k$ are mutually orthogonal.

Property 1 means that if $\psi$ satisfies the equation $\mathcal H\psi = \epsilon\psi$ then $\psi$ can be expressed as a combination (or superposition) of $\psi_k$s: $$ \psi(x) = \sum_k A_k\psi_k(x). $$ And property 2 means that if $k'\neq k+G$ then their inner product is zero: $$ \int_{-\infty}^\infty \psi_k(x)^* \psi_{k'}(x)\,\mathrm dx = 0. $$

The second property has some implications in quantum mechanics. In principle, we could measure the wavevector of any wavefunction $\psi$. If $\psi$ is one of our special solutions, $\psi_k$, then measuring the wavevector is guaranteed to yield $k$. The same cannot be said for an arbitrary $\psi$. For example, if we have wavevectors $k_1$ through $k_4$, then we can construct a wavefunction $\psi$: $$ \psi(x) = \frac{1}{2}\left(\psi_{k_1}(x) + \psi_{k_2}(x) + \psi_{k_3}(x) + \psi_{k_4}(x)\right) $$ If we try to measure the wavevector of $\psi$, there is a $1/4$ chance of measuring each $k_1$ through $k_4$.

If I'm interpreting correctly, this is the troublesome sentence:

If one particular wavevector $k$ is contained in a $\psi$, then all other wavevectors in the Fourier expansion of this $\psi$ will have the form $k+G$, where $G$ is any reciprocal lattice vector.

Now, you ask why. The answer is that, well, Charles Kittel is lying here. They're not looking for just any solutions, they're looking for a helpful set of solutions. Let's take a step back and derive the central equation $$ (\lambda_k - \epsilon)C(k) + \sum_{G}U_GC(k-G) = 0. $$

Starting at the previous equation: $$ \sum_k \lambda_k C(k) e^{ikx} + \sum_G\sum_k U_GC(k)e^{i(k+G)x} = \epsilon\sum_kC(k)e^{ikx} $$ The second term on the right side can be rearranged: $$ \sum_G\sum_k U_GC(k)e^{i(k+G)x} = \sum_k\left(\sum_GU_gC(k-G)\right)e^{ikx} $$ and then we get to rearrange the previous equation to have a sum over $k$ on one side, zero on the other: $$ \sum_k \left((\lambda_k - \epsilon) C(k) + \sum_GU_gC(k-G)\right)e^{ikx} = 0 $$ Only way for this to be true is for the coefficients to each be zero: $$ (\lambda_k - \epsilon) C(k) + \sum_GU_gC(k-G) = 0 $$ Once we have a $k_0$ value with $C(k_0)\neq 0$, we have all of the $C(k_0-G)$ coefficients determined by the central equation. Moreover, saying that $C(k\neq k_0-G) = 0$ will yield a valid wavefunction with energy $\epsilon$.

But what if $C(k\neq k_0-G)\neq 0$? Well, the honest answer is that we end up with the same result: a wavefunction with energy $\epsilon$. There's nothing actually stopping us from having a wavefunction which includes all wavevectors $k$ with the right energy $\epsilon$.

But the helpful part of using $C(k\neq k_0-G) = 0$ is that it organizes the solutions into a set of $\psi_k$ with wonderful properties:

1. The solutions $\psi_k$ form a basis.
2. The solutions $\psi_k$ are mutually orthogonal.

Property 1 means that if $\psi$ satisfies the equation $\mathcal H\psi = \epsilon\psi$ then $\psi$ can be expressed as a combination (or superposition) of $\psi_k$s: $$ \psi(x) = \sum_k A_k\psi_k(x). $$ And property 2 means that if $k'\neq k+G$ then their inner product is zero: $$ \int_{-\infty}^\infty \psi_k(x)^* \psi_{k'}(x)\,\mathrm dx = 0. $$

The second property has some implications in quantum mechanics. In principle, we could measure the wavevector of any wavefunction $\psi$. If $\psi$ is one of our special solutions, $\psi_k$, then measuring the wavevector is guaranteed to yield $k$. The same cannot be said for an arbitrary $\psi$. For example, if we have wavevectors $k_1$ through $k_4$, then we can construct a wavefunction $\psi$: $$ \psi(x) = \frac{1}{2}\left(\psi_{k_1}(x) + \psi_{k_2}(x) + \psi_{k_3}(x) + \psi_{k_4}(x)\right) $$ If we try to measure the wavevector of $\psi$, there is a $1/4$ chance of measuring each $k_1$ through $k_4$.