Proof that the real vector space of $C^\infty$ functions with $f''(x) + f(x) = 0$ is two-dimensional
I shall prove a general statement. Let $p(t)\in\mathbb{C}[t]$ be a monic polynomial of degree $n\in\mathbb{Z}_{>0}$ and suppose that $V:=\mathcal{C}^\infty(\Omega,\mathbb{C})$ is the $\mathbb{C}$-vector space of complex-valued functions on $\Omega$, where $\Omega$ is a nonempty connected open set of $\mathbb{R}$. Write $D:V\to V$ for the differentiation map: $$\big(D(f)\big)(x):=f'(x)=\left.\left(\frac{\text{d}}{\text{d}u}f(u)\right)\right|_{u=x}$$ for all $x\in \Omega$. Then, $\ker\big(p(D)\big)$ is an $n$-dimensional $\mathbb{C}$-vector subspace of $V$. (Even more generally, if $\Omega$ is an open set of $\mathbb{R}$ with $c$ connected components, then $\ker\big(p(D)\big)$ is $cn$-dimensional over $\mathbb{C}$. Also, $V$ can be replaced by the space of holomorphic functions on $\Omega$, if $\Omega$ is taken to be an open subset of $\mathbb{C}$.)
First, write $p(t)=\prod_{i=1}^k\,\left(t-z_i\right)^{m_i}$, where $z_1,z_2,\ldots,z_k\in\mathbb{C}$ are pairwise distinct, and $m_1,m_2,\ldots,m_k\in\mathbb{Z}_{>0}$. Define $$p_i(t):=\left(t-z_i\right)^{m_i}\text{ and }q_i(t):=\frac{p(t)}{p_i(t)}\,.$$ Then, the fact that $$p(t)=p_i(t)\,q_i(t)\text{ and }p_i(t)\,f_i(t)+q_i(t)\,g_i(t)=1$$ for some $f_i(t),g_i(t)\in\mathbb{C}[t]$ implies that $$\ker\big(p(D)\big)=\ker\big(p_i(D)\big)\oplus \ker\big(q_i(D)\big)\,.$$ By induction, we see that $$\ker\big(p(D)\big)=\bigoplus_{i=1}^k\,\ker\big(p_i(D)\big)\,.$$
Thus, it boils down to studying $\ker\big(p(D)\big)$, where $p(t)=(t-z)^m$ for some $z\in\mathbb{C}$ and $m\in\mathbb{Z}_{>0}$. However, consider the map $M_z:V\to V$ given by $$\big(M_z(f)\big)(x):=\exp(zx)\,f(x)$$ for all $x\in \Omega$. As $$p(D)=M_z\,D^m\,M_{-z}=M_z\,D^m\,\left(M_z\right)^{-1}\,,$$ $p(D)$ and $D^m$ are conjugate linear maps. Therefore, $$\ker\big(p(D)\big)=M_z\big(\ker\left(D^m\right)\big)\,.$$ Since $\ker\left(D^m\right)$ is $m$-dimensional and $M_z$ is a vector-space automorphism, $$\dim\Big(\ker\big(p(D)\big)\Big)=\dim\Big(\ker\big(D^m\big)\Big)=m\,.$$ In fact, $\ker\big(p(D)\big)$ consists of elements of the form $M_z(f)$, where $f:\Omega\to\mathbb{C}$ is a polynomial function of degree less than $m$.
Alternatively, let $U:=\ker\big(p(D)\big)$. Then, show that $p(t)$ is the minimal polynomial of $D|_U:U\to U$. From my post here, $U$ decomposes as $$U=\bigoplus_{i=1}^m\,\ker\left(\big(D-z_i\,\text{id}_U\big)^{m_i}\right)\,,$$ if $p(t)=\prod_{i=1}^k\,\left(t-z_i\right)^{m_i}$.
Let $\mathcal{V}$ denote the $\mathbb{C}$-vector space of $\mathbb{C}$-valued sequences $\left(X_N\right)_{N\in\mathbb{Z}_{\geq 0}}$. Fix $a_0,a_1,\ldots,a_{n-1}\in\mathbb{C}$. The same technique can be employed to show that the solutions $\left(X_N\right)_{N\in\mathbb{Z}_{\geq 0}}\in\mathcal{V}$ of a recurrence relation $$X_{N+n}+a_{n-1}\,X_{N+n-1}+\ldots+a_1\,X_{N+1}+a_0\,X_N=0$$ for all $N\in\mathbb{Z}_{\geq 0}$ form an $n$-dimensional $\mathbb{C}$-vector subspace of $\mathcal{V}$.
P.S.: If you want to consider $V_\mathbb{R}:=\mathcal{C}^\infty(\Omega,\mathbb{R})$ instead, then use the fact that $V=\mathbb{C}\underset{\mathbb{R}}{\otimes} V_{\mathbb{R}}$. That is, if $p(t)\in\mathbb{R}[t]$, then the kernel $K_\mathbb{R}$ of $\big.p(D)\big|_{V_\mathbb{R}}:V_\mathbb{R}\to V_\mathbb{R}$ satisfies $$\mathbb{C}\underset{\mathbb{R}}{\otimes} K_\mathbb{R}=\ker\big(p(D)\big)\,.$$
HINT: Show that the function $$ \mathbb{R} \ni t \mapsto \left (\matrix {\cos t & - \sin t \\ \sin t & \cos t } \right )\cdot \left ( \matrix{ f(t) \\ f'(t) } \right ) \in \mathbb{R}^2$$ is constant
$\bf{Added:}$
Let's show that if $n$ solutions of the linear equation $y'(t) = A(t) y(t)$ have an invertible Wronskian $W(t)$, and $f$ is another solution then $f(t)$ is a linear combination of these solutions. Recall that the columns of $W$ are these $n$ solutions, so we have $W'(t) = A(t) W(t)$.
Indeed, consider the function $$t\mapsto W(t)^{-1} f(t)$$ Its derivative equals $$(W(t)^{-1})' f(t) + W(t)^{-1} f'(t)$$ Now we have $$(W(t)^{-1})'= - W(t)^{-1} W(t)' W(t)^{-1} $$ We get $$(W(t)^{-1})' f(t) + W(t)^{-1} f'(t) = \\=- W(t)^{-1} A(t) W(t) W(t)^{-1} f(t) + W(t)^{-1} A(t) f(t)= \\ =- W(t)^{-1} A(t) f(t) + W(t)^{-1} A(t) f(t) = 0$$
Hence the function $$ W(t)^{-1} f(t) $$ is a constant $c$ (in $\mathbb{R}^n)$ and so $f(t) = W(t) \cdot c$ is a linear combination of the columns of $W(t)$.
Read about the Wronskian.
For three supposedly linearly independent $C^{\infty}$ solutions $f$ $g$, and $h$, the Wronskian determinant should be nonzero. But since the last row will be $-1$ times the first row, the Wronskian will be $0$. This contradicts the possibility for three independent solutions.