Non-compactness of operator

Question

Consider a linear operator $T$ on the Hilbert Space $l^2$ given by $$T(x_1,x_2,...)=(a_0x_1,a_1x_2,\cdots),$$
where the sequence $(a_n)$ is dense in $[0,1]$. I deduced that $T$ cannot be compact because there will exist a subsequence of $(a_n)$ which will converge to 1 and hence the operator will ultimately become the identity operator on $l^2$ which is not compact. Can someone please verify if my reasoning is right?