I think I get part a, but not sure how to start part b. Thanks in advance for the help.
a) Show that for any A ∈ $R^{m×n}$, it holds that $A^T A$ is positive semidefinite: I think this follows with $x^T A^TAX = (Ax)^T (Ax) = ||Ax||^2_2 \geq 0$
b) Show that for any symmetric B ∈ $R^{n×n}$ there is a real number c such that cI + B is positive definite
Your proof for (a) is correct. (b) is quite easy if you have spectral theory developed already. Since $B$ is symmetric, it can be orthogonally diagonalized $B = Q^T \Lambda Q$ where $\Lambda = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)$. Take $c := \max_i |\lambda_i| + 1$. Then
$$ B + cI = Q^T\Lambda Q + Q^TcI Q = Q^T (\Lambda + cI) Q. $$
By construction, $\Lambda + cI$ has all positive elements on its diagonal so $B+cI$ is positive definite since all its eigenvalues are positive.
Without spectral theory, we have to work a slight bit more. Consider the function that takes every vector $x \in \mathbb{R}^n$ to $x^T B x$. You, with some work, can show this function is continuous. Now consider only vectors $x$ with unit length, $\|x\| = 1$. Then by a very powerful tool in mathematical analysis called compactness, we know that $x^T B x$ is bounded. That is, there exist constants $C_1$ and $C_2$ such that $C_1 \le x^T B x \le C_2$. (If all this about continuity and compactness doesn't help, just think of all this as a formal mathematical way of saying matrices can't stretch input vectors infinitely much.)
Set $c := |C_1| + 1$. Then for all unit vectors $x$,
$$ x^T (cI + B) x = c \underbrace{x^T x}_{=1} + \underbrace{x^T B x}_{\ge C_1} \ge c + C_1 \ge 1> 0. $$
For any vector $v \ne 0$, we have $v = \|v \| (v /\|v\|)$. Then $v/\|v\|$ is a unit vector so
$$ v^T (cI + B) v = \|v \|^2 \left(\frac{v}{\|v\|}\right)^T (cI + B) \left(\frac{v}{\|v\|}\right) \ge \|v\|^2 \cdot 1 > 0. $$
So $cI + B$ is positive definite.