An nth-order ODE has n linearly independent solutions

Your proof is now good. For the homogeneous non-constant coefficient system $$a_n(t)x^{(n)}+\cdots +a_0(t)x=0$$ the solutions again form a vector space, there is nothing different in this respect. How to find its dimension (in particular, prove that the dimension is not zero)?

One cannonball way to proceed is to rewrite it as a first order vector valued equation by introducing variables $x_1=x',\cdots$. This furnishes the first order equation $$X'(t)=A(t)X(t).$$ Here the function $A$ is assumed nicely behaved from some interval $(a,b)$ to the Banach space $\mathbb{R}^n$, e.g. take the entries to be Lipschitz and $a_n(t)\neq 0$ on $(a,b)$ to avoid the degenerate locus (in fact, assume $a_i/a_n(t)$ are also Lipschitz on the interval). Then we can form any IVP we want $$X'=A(t)X(t),\quad X(0)=X_0$$ and the Picard-Lindelof theorem furnishes a unique solution with the initial condition $X_0$ (if one tries to simplify the proof of P-L for linear systems one still ends up having to perform the crucial approximation process and prove the uniform convergence, which is the big deal in P-L itself).

But now the fact that you can find solutions corresponding to $n$ linearly independent choices of $X_0$ implies that the vector space of solutions is exactly $n$-dimensional by the theorem on the Wronskian (the Wronskian of $n$ solutions is either identically zero or never zero; if the initial vectors $X_0$ are linearly independent, the Wronskian is nonzero somewhere, so it is never zero on $(a,b)$ and thus the solutions are independent).