The map that sends $A$ to its greatest eigenvalue is continuous.

On a normed space, the norm map $v \in V \mapsto \|v\| \in \mathbb R$ is continuous.

The spectral norm of a matrix $A$ is given by its largest singular value, which is also the square root of the largest eigenvalue of $AA^*$.

For real symmetric matrices, $AA^*=A^2$ and so the spectral norm of $A$ is largest eigenvalue of $A$.

Thus the map $A \mapsto \|A\|_2=\lambda_{max}(A)$ is continuous.

Hint: You can use Rouché's Theorem in complex analysis to prove that the (complex) roots of a complex polynomial vary continuously with the coefficients of the polynomial.