First, let us see that all eigenvalues have either positive or negative real part. Let $(\mu,\nu)$ be the standard basis for $\mathbb{R}^{2}$. Decompose every vector $\xi\in\mathbb{R}^{2}$ into $\xi=\xi_{\mu}\mu+\lambda\nu$. We consider the matrix-valued quadratic bilinear form $P:\mathbb{R}^{2}\rightarrow\mathrm{Hom}{\mathbb(\mathbb{R}^{d})}$ given by 
$$P(\xi_{\mu}\mu+\lambda\nu)=\lambda^{2}A-\lambda\xi_{\mu} B+\xi_{\mu}^{2}C$$
Note how for every $\xi\in\mathbb{R}^{2}$, by the assumptions in the question we find 
$$|\xi|^{2}I\leq P(\xi)\leq 2|\xi|^{2}I.$$
In particular, $P(\xi)$ is nonsingular for all $\xi\ne 0$.

By formally allowing $\lambda,\xi_{\mu}$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y_{0}e^{\lambda t}$,
$$P(\mu+i\lambda\nu)y_{0}=0$$
For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads,
$$P(\mu-\eta\nu)y_{0}=0$$
which implies $y_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part. 

This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result.

Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the  fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. For all $\eta\in\mathbb{R}$, define $Q(\eta)$ to be the operator
$$Q(\eta)=P(\mu-\eta\nu).$$ 

Decompose $Q(\eta)=M(\eta)L(\eta)$, where $L$ and $M$ are both linear in $\eta$.

By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$, 
the ODE can formally be written as,
$$Q(i\frac{d}{dt})y(t)=P(\mu-i\frac{d}{dt})y(t)=0$$
Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain,
$$0=(Q(i\frac{d}{dt})y,y)_{L^{2}(\mathbb{R^+})}=(L(i\frac{d}{dt})y,M^{*}(i\frac{d}{dt})y)_{L^{2}(\mathbb{R^+})}$$
By the Plancheral theorem, this quantity becomes,
$$\int_{-\infty}^{\infty}(P(\mu-\eta\nu)\hat{y}(\eta),\hat{y}(\eta))_{\mathbb{R}^{d}}d\eta=0$$.  
And since $P(\xi)\geq |\xi|^{2}I$ for all $\xi\ne 0$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described. 

The proof also shows that all fundamental solutions must be of the form $y(t)=y_{0}e^{\lambda t}$, because if $y(t)=y_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.