Convexity of $\{x\in \mathbb R^n,\;x^TAx\leq t\}$ for all $t$ implies that $A$ is p.s.d.

You don't need to consider two different eigenvalues. Let $\lambda_\min$ be the minimum eigenvalue of $A$ and $u$ be a corresponding unit eigenvector. Then $\{u,-u\}\subset C=\{x: x^TAx\le\lambda_\min\}$. However, by assumption, $C$ is convex. Therefore $x=\frac{u+(-u)}{2}=0\in C$, meaning that $0\le\lambda_\min$. Hence $A$ is positive semidefinite.

Suppose $y^TAy < 0$ and $L=\{x | x^T Ax \le y^TAy \}$ is convex. Since $(-y)^TA(-y) = y^TAy < 0$ then ${1 \over 2} (y+(-y)) = 0 \in L$, but this is a contradiction hence $y^T A y \ge 0$ for all $y$.