Calculate $\begin{Vmatrix}1&2\\2&4\end{Vmatrix}$

You have shown that $$\max \dfrac{\Vert Ax \Vert}{\Vert x \Vert} \leq 5$$ To show that equality is attained, choose $x = \begin{bmatrix} 1\\ 2\end{bmatrix}$. We have $$Ax = \begin{bmatrix} 1 & 2\\ 2 & 4\end{bmatrix} \begin{bmatrix}1\\2 \end{bmatrix} = \begin{bmatrix} 5 \\ 10\end{bmatrix} = 5x$$ Hence, $$\dfrac{\Vert Ax \Vert}{\Vert x \Vert} = 5$$

In general, note that if $A = uu^T$, then the reduced SVD of $A$ is $$A = \Vert u \Vert_2^2 \left(\dfrac{u}{\Vert u \Vert_2} \right) \left(\dfrac{u}{\Vert u \Vert_2} \right)^T$$ Hence, $\Vert A \Vert_2 = \Vert u \Vert_2^2$. In your case, $$A = \begin{bmatrix} 1\\ 2\end{bmatrix} \begin{bmatrix} 1 & 2\end{bmatrix}$$ Hence, $u = \begin{bmatrix} 1\\ 2\end{bmatrix}$ and thereby $$\Vert A \Vert_2 = \Vert u \Vert_2^2 = 1^2 + 2^2 = 5$$

Consider the vector (1,2) again...