If $A = R R^T$, prove that $||R_1 R_1^T||_2 \le ||A||_2$ where $R_1$ is first column of $R$.

Hint: There's no need to invoke the SVD here. Note that $$ A = RR^T = \sum_{i=1}^n R_iR_i^T $$ is a sum of positive semidefinite matrices. If $A,B$ are positive semidefinite, then there is an inequality relating the eigenvalues of $A$ and $B$ to the eigenvalues of $A + B$ that I suspect you have somewhere in your notes. If you do not, then you can deduce this inequality using the fact that for symmetric matrices $M$, $$ \lambda_{\max}(M) = \max_{\|x\| = 1} x^TMx. $$

To elaborate, we have the following.

$$ \lambda_{\max}(A + B) = \max_{\|x\| = 1} x^T(A + B)x = \max_{\|x\| = 1}(x^TAx + x^TBx) \\ \geq \max_{\|x\| = 1} (x^TAx + 0) = \lambda_{\max}(A). $$ From there, set $A = R_1R_1^T$ and $B = \sum_{i=2}^n R_iR_i^T$.