Let $A$ and $B$ be $n \times n$ positive definite matrices. Let $C$ be any $m \times n$ matrix. Can we always find $\alpha$ and $\beta$ such that the matrix
$$\begin{bmatrix} A + \alpha C^\top C + \frac{1}{\beta}I & - \frac{1}{\beta} I \\ - \frac{1}{\beta} I & -B + \frac{1}{\beta}I \end{bmatrix}$$
is positive definite?
The existence can be proven in different ways. We note first that we need $\beta>0$.
A possible solution relies on the consideration of a perturbation argument on the matrix $\beta M$ where $M$ is the matrix of interest. When $\beta=0$, we get that
$$M\beta\vert_{\beta=0}=\begin{bmatrix}I & -I\\-I & I\end{bmatrix}.$$
This matrix has $n$ eigenvalues at zero, and $n$ eigenvalues at 2. Let us denote those zero eigenvalues by $\lambda_1(M\beta\vert_{\beta=0}), \ldots, \lambda_n(M\beta\vert_{\beta=0})$; i.e. the first $n$ eigenvalues of $M\beta\vert_{\beta=0}$.
So, we need to look at what happens to the zero eigenvalues when we slightly increase $\beta$ to positive values. The zero eigenvalues are semi-simple and their normalized right eigenvectors are given by
$$u=\dfrac{1}{\sqrt{2}}\begin{bmatrix}I\\I\end{bmatrix}.$$
Since the matrix is symmetric, then the normalized left eigenvectors are simply given by $u^T$.
It is known that semi-simple eigenvalues are locally differentiable and that they change according to
$$\lambda_i(\beta M)=\lambda_i(M\beta\vert_{\beta=0})+\beta \lambda_i\left(u^T\begin{bmatrix} A + \alpha C^\top C& 0\\ 0 & -B \end{bmatrix}u\right)+o(\beta),$$
for $i=1,\ldots,n$. That is, the direction of change of those zero eigenvalues is equal to the eigenvalues of the matrix in brackets (which can be understood here as a perturbation matrix projected on the eigenspace associated with the zero eigenvalues of the matrix $\beta M$ evaluated at zero).
Evaluating term in brackets yields $\frac{1}{2}(A+\alpha C^TC-B)$ which shows that if this matrix is positive definite, then the eignvalues of $\beta M$ will all be positive for some sufficiently small $\beta>0$.
Finally, the condition that there exists an $\alpha$ such that $A+\alpha C^TC-B$ is positive definite is equivalent to saying that the matrix $N^T(A-B)N$ is positive definite where $N$ is a basis for the right null-space of the matrix $C$. This is a direct consequence of the Finsler's lemma.
It is interesting to check whether any of those condition is necessary. A necessary condition is that $\beta>0$ and $-\beta B+I$ is positive definite. This is equivalent to saying that $\beta\in(0,1/\lambda_{\mathrm{min}}(B))$.
Another necessary condition is that $A-B+\alpha C^TC$ is positive definite. Indeed, we have that
$$\begin{bmatrix} I & I\\ I & -I \end{bmatrix}\begin{bmatrix} A + \alpha C^\top C + \frac{1}{\beta}I & - \frac{1}{\beta} I \\ - \frac{1}{\beta} I & -B + \frac{1}{\beta}I \end{bmatrix}\begin{bmatrix} I & I\\ I & -I \end{bmatrix}=\begin{bmatrix} A + \alpha C^\top C -B & A + \alpha C^\top C+B\\ A + \alpha C^\top C+B & A + \alpha C^\top C+B+4I/\beta \end{bmatrix}. $$
This means that $M$ is positive definite if and only if the matrix on the right hand side is positive definite. A necessary condition for that is that the diagonal blocks be positive definite as well. This proves the necessity of the positive definiteness of $A + \alpha C^\top C -B$.