How does one prove that the spectral norm is less than or equal to the Frobenius norm?

The Frobenius norm of $A$ is the squareroot of the sum of all of the eigenvalues (the trace) of $A^*A$, and since all eigenvalues of $A^*A$ are nonnegative, it follows that the largest eigenvalue is less than or equal to the sum of all of the eigenvalues.

If $p=2$ it's the Frobenius norm, right?

No, if $p=2$ that is another characterization of the spectral norm. A proof of this is sketched in Michalis's answer.

Any matrix norm induced by a norm on your vector space (over $\mathbb{R}$ or $\mathbb{C}$) that also satisfies $\|A^{*}\|=\|A\|$ will be greater than or equal to the spectral norm. Let $\lambda$ denote the largest singular value of A (the square root of the largest eigenvalue of ($A^*A$) ) and $v$ the corresponding eigenvector. Let $\|A\|$ denote the matrix norm induced by a norm on the vector space: $$ \|A\|^2=\|A^{*}\|\cdot\|A\|\geq\|A^{*}A\|=\max\frac{\|A^{*}Ax\|}{\|x\|}\geq\frac{\|A^{*}Av\|}{\|v\|}=\lambda $$ and so $\|A\|\geq\sqrt{\lambda}$

For the 2-norm you actually have equality, which you can show by singular value decomposition. We can take an orthonormal basis of eigenvectors for $A^*A$ (with respect to the usual scalar product that also induces the 2-norm). Denote this basis by $v_1,\ldots,v_n$ with eigenvalues $\lambda=\lambda_1,\ldots,\lambda_n$. For any vector $x=\sum x_i v_i$ we have $$ \|Ax\|_2^2=\overline{x}^TA^{*}Ax\leq\overline{x}^T\sum\lambda_i x_i v_i=\sum\lambda_i |x_i|^2\leq \lambda \|x\|_2^2 $$ So $\|A\|_2\leq \sqrt{\lambda}$ and both inequalities together show $\|A\|_2=\sqrt{\lambda}$.