We have $A \in \mathbb{R}^{n \times n}$ which is symmetric and positive-definite. Also, $A$ is a block matrix:
$$A = \begin{pmatrix} A_{11} & A_{21}^{\top} \\ A_{21} & A_{22} \\ \end{pmatrix}.$$
We can then obtain a block Cholesky factorization,
$$A = \begin{pmatrix} L_{11} & 0 \\ L_{21} &L_{22} \\ \end{pmatrix}\begin{pmatrix} L_{11}^{\top} & L_{21}^{\top} \\0 &L_{22}^{\top} \\ \end{pmatrix}.$$
Is it true that, $$\|L_{22}\|_{2} \le \|A\|_{2}?$$
I don't think it is.
I assume that $\|A\|_2$ refers to the spectral norm of $A$ (the maximal singular value of $A$).
Your statement does not hold. As an example, consider $$ A = \pmatrix{1/4 & 0\\0 & 1/4}, \quad L = \pmatrix{1/2 & 0\\0 & 1/2}. $$
It is generally true that $\|L_{22}\|_2 \leq \sqrt{\|A\|_2}$. We can see this as follows. Denote $$ L = \pmatrix{L_{11} & 0\\L_{21} & L_{22}}. $$ We will use the following two facts:
With that, we note that \begin{align} \sqrt{\|A\|_2} &= \|L\|_2 = \max \{x^TLy: \|x\| = \|y\| = 1\} \\ & = \max \Bigg\{ \pmatrix{x_1\\ x_2}^T \pmatrix{L_{11} & 0\\ L_{21} & L_{22}}\pmatrix{y_1\\y_2}: \\ & \qquad \qquad \|x_1\|^2 + \|x_2\|^2 = \|y_1\|^2 + \|y_2\|^2 = 1\Bigg\} \\ & \geq \max \left\{\pmatrix{0\\ x_2}^T \pmatrix{L_{11} & 0\\ L_{21} & L_{22}}\pmatrix{0\\y_2}:\|x_2\|^2 = \|y_2\|^2 = 1\right\} \\ & = \max \{x_2^TL_{22}y_2 : \|x_2\| = \|y_2\| = 1\} = \|L_{22}\|. \end{align}