Prove $\|A\| \leq \|A\|_{HS}$, where $\|A\|$ is the operator norm of A I Am trying to solve the following problem Let
$H_1
,H_
2$
be  Hilbert  spaces.   Let
$A
\in B(H_1
,H_2
)$  be  a  Hilbert-Schmidt  operator.   For  a  complete
orthonormal sequence $(
u_n
)$ in
$H_1$, define the
Hilbert-Schmidt norm
$\|.\|_{HS}$
by 
$\|A\|_{HS}=\left(\sum^\infty_{n=1}\|A(u_n)\|^2\right)^\frac{1}{2} $
Prove  $\|A\| \leq \|A\|_{HS}$, where $\|A\|$ is the operator norm of A.
My current attempt is as follows:
Let $x \in H_1$ such that $ x=\sum^\infty_{n=1} \langle x,u_n\rangle u_N$ and $\|x\|=\left(\sum^\infty_{n=1}|\langle x,u_n\rangle|^2\right)^\frac{1}{2} $. Then we have, 
$\|A(x)\|=\|A\left(\sum^\infty_{n=1} \langle x,u_n\rangle u_N\right)\| =\|\sum^\infty_{n=1}A\left( \langle x,u_n\rangle u_n\right)\| = \|\sum^\infty_{n=1} \langle x,u_n\rangle A\left(u_n\right)\| \leq \|\sum^\infty_{n=1}A(u_n)\|\ \|\sum^\infty_{n=1}\langle x,u_n\rangle\|$ 
I am unsure where to go from here and feel I have made a mistake? Thank you in advance.

In the last inequalities you have used an inequality of the form:
$$\left\Vert\sum_n a _n b_n \right\Vert \leq  \left\Vert\sum_n a _n \right\Vert \left|\sum_n  b_n \right|$$
which is not true. However you can use a Cauchy-Schwartz inequality which gives:
$$\sum_n \left\Vert a _n\right\Vert |b_n|  \leq  \left(\sum_n \left\Vert a _n \right\Vert^2  \right)^\frac{1}{2} \left(\sum_n \left| b _n \right|^2  \right)^\frac{1}{2} $$
