If $A^*x_n \to y$, there exists a sequence such that $A^*Ay_n \to y$. I'm struggling with a problem from Young's Introduction to Hilbert Space (7.30 to be more specific)

Let $\mathbb{H}$ be a Hilbert space, $A\in B(\mathbb{H})$ (ie a bounded operator) and $(x_n)_{n=1}^{\infty}$ a sequence in $\mathbb{H}$.
Prove that if $A^*x_n\rightarrow y$, there exists a sequence $(y_n)_{n=1}^{\infty}$ such that $A^*Ay_n \rightarrow y$.

I'm 99% sure this has something to do with the relationship between the kernel and the image of $A$ and $A^*$ ($\ker A^*$ being the orthogonal of the image of $A$, etc), but I haven't been able to make much progress. Could I get a hint on how to proceed?
Recall that $\ker A = \ker A^*A$. Namely, clearly $\ker A \subseteq \ker A^*A$. Conversely

$$x \in \ker A^*A \implies A^*Ax = 0 \implies 0 = \langle A^*Ax, x\rangle = \langle Ax, Ax\rangle \implies Ax = 0\implies x \in \ker A$$

Since $(\ker T)^\perp  = \overline{\operatorname{Im} T^*}$, taking orthogonal complements in the above relation gives

$$\overline{\operatorname{Im}A^*} = \overline{\operatorname{Im}A^*A}$$

Therefore, $A^*x_n \to y$ implies that $y \in \overline{\operatorname{Im}A^*}$, so $y \in \overline{\operatorname{Im}A^*A}$ implying that there exists a sequence $(y_n)$ such that $A^*Ay_n \to y$.
The image of a linear operator is closed, so $y$ is a limit of elements of an image iff it is an element of that image.  So this is equivalent to showing that given any $v$, there is some $u$ such that $A^* Au = A^*v$. 

$v$ can be written as $h+p$, where $h$ is in $ker A^*$ and $p$ is orthogonal to $ker A^*$, and $A^*v =A^*p$. So we need, given arbitrary $p$, to find $u$ such that $Au = p$.