By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that \begin{equation} X=\left[ \begin{array}{cc} x & y \\ y & z \end{array} \right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge0, \quad xz\ge y^2,\tag{1} \end{equation} where $$ S_+^2 = \left\{X\in S^2 | X \succeq 0\right\}. $$
I have failed to prove the $(1)$.
$$ \begin{align} \forall \alpha,\beta,\quad \left[ \begin{array}{cc} \alpha & \beta \end{array} \right] \left[ \begin{array}{cc} x & y \\ y & z \end{array} \right] \left[ \begin{array}{cc} \alpha \\ \beta \end{array} \right] &\ge 0\\ \alpha^2x + 2\alpha\beta y + \beta^2z &\ge 0\\ \left(\alpha\sqrt{x} + \beta\sqrt{z}\right)^2 + 2\alpha\beta(y-\sqrt{xz}) &\ge 0 \\\therefore \alpha\beta(y-\sqrt{xz})\ge0\\ \text{if}\quad\alpha\beta\ge0,\quad \text{then} \quad y^2\ge xz\\ \text{if}\quad\alpha\beta\le0,\quad \text{then} \quad y^2\le xz \end{align} $$
How can I reach to $$ x\ge0,\quad z\ge0,\quad xz\ge y^2\qquad? $$
This is essentially the implication from the Sylvester's criterion extended to the case of positive semi-definite matrices.
All of the leading principal minors must be nonnegative. So you immediately get $x\ge 0, x z\ge y^2, z\ge 0$.
Following your proof for $2\times2$ matrices, take consequently $\alpha = 0$ and $\beta = 0$ to get $x\ge 0, z\ge 0$. The last step is to take $\alpha = \beta = 1$
$$2y \ge -x-z\Rightarrow 4y^2 \ge (x+z)^2 \ge 4xz$$