Help proving orthogonal compliment of nullspace of adjoint operator is the closure of the range I'm supposed to show that $N(A^{*})^{\perp}=\overline{A(X)}$.
I've already shown that $N(A^{*})=A(X)^{\perp}$, and I've used that in my proof. I feel like my proof is not complete. Some of the statements I make feel a little vague, nor am I not completely sure whether they are true. As extra information, we're dealing with a linear and bounded operator $A:X\rightarrow Y$, where both $X$ and $Y$ are Hilbert spaces.
Here's my proof:
One can notice that $\overline{A(X)}\subseteq (A(X)^{\perp})^{\perp}$. This is because if we consider an arbitrary $y\in \overline{A(X)}$, then this means there exists a sequence $\{x_n\}\subset X$ such that $\underset{n\rightarrow\infty}{\lim}Ax_n=y$. Now, consider an arbitrary element $v\in A(X)^{\perp}$, then we have
$$\langle y,v\rangle=\langle\underset{n\rightarrow\infty}{\lim}Ax_n,v\rangle=\underset{n\rightarrow\infty}{\lim}\langle Ax_n,v\rangle=0$$
Hence we can see that $y\in (A(X)^{\perp})^{\perp}$, which implies $\overline{A(X)}\subseteq (A(X)^{\perp})^{\perp}$.
Now, consider some arbitrary $\varphi\in (A(X)^{\perp})^{\perp}$. This means that for all $\psi\in A(X)^{\perp}$ then we have $\langle\varphi,\psi\rangle=0$. Now define the orthogonal projection operator $P:Y\rightarrow\overline{A(X)}$. Since $P\varphi\in\overline{A(X)}\subseteq (A(X)^{\perp})^{\perp}$ that implies $P\varphi\in (A(X)^{\perp})^{\perp}$. Now consider for $v\in A(X)^{\perp}$
$$\langle\varphi-P\varphi,v\rangle=\langle\varphi,v\rangle-\langle P\varphi,v\rangle=0$$
since both $\varphi,P\varphi\in (A(X)^{\perp})^{\perp}$. This shows that $\varphi-P\varphi\perp A(X)^{\perp}$. Furthermore, it can be seen that $\varphi-P\varphi\perp A(X)$, since we are subtracting all components of $\varphi$ that are projected onto $\overline{A(X)}$. this means that $\varphi-P\varphi\perp A(X)$ and $\varphi-P\varphi\perp A(X)^{\perp}$, thus $\varphi-P\varphi$ must be in the closure, i.e., $\varphi-P\varphi\in\overline{A(X)}$. But since we've subtracted all the projections of $\varphi$ onto $\overline{A(X)}$, then $\varphi-P\varphi$ can only be in the closure if $\varphi=P\varphi\in\overline{A(X)}$. From this we have $(A(X)^{\perp})^{\perp}\subseteq\overline{A(X)}$, hence yielding the equality $(A(X)^{\perp})^{\perp}=\overline{A(X)}$. By use of the previously proved statement, we have that $A(X)^{\perp}=N(A^{*})$ thus $N(A^{*})^{\perp}=\overline{A(X)}$, as desired.
If you can spot mistakes or something I don't make clear, then please let me know.
Thanks in advance.
 A: Since you have already shown that $N(A^{*})=A(X)^{\perp}$ (also proved here) then by taking the orthogonal complement on both sides you simply need to show that $$\overline{A(X)}= {A(X)^{\perp}}^{\perp}.$$
As alluded to in comments, it is well known that a closed subspace of a Hilbert space is equal its double orthogonal complement (for example Corollary 1.5 here), so $$\overline{A(X)}= {\overline{A(X)}^{\perp}}^{\perp}.$$
What's left is to show that $${\overline{A(X)}^{\perp}}^{\perp}={A(X)^{\perp}}^{\perp}.$$
Since $A(X)\subset \overline{A(X)}$ it is easy to see that  $\overline{A(X)}^{\perp} \subset {A(X)}^{\perp}$. Conversely, if $z\in {A(X)}^{\perp}$ then for arbitrary $w\in \overline{A(X)}$ we have that $w=\underset{n\rightarrow\infty}{\lim} w_n$ for some $\{w_n\}\subset A(X)$, so $$\langle w , z \rangle=\langle \underset{n\rightarrow\infty}{\lim} w_n , z \rangle = \underset{n\rightarrow\infty}{\lim}\langle  w_n , z \rangle=\underset{n\rightarrow\infty}{\lim} 0 = 0$$
which proves that ${A(X)}^{\perp}\subset \overline{A(X)}^{\perp}$ and hence  ${A(X)}^{\perp}= \overline{A(X)}^{\perp}$. Taking orthogonal complements on both sides gives the required  fact that ${\overline{A(X)}^{\perp}}^{\perp}={A(X)^{\perp}}^{\perp}.$

Specifically relating to your proof, you could clarify what you did after defining the orthogonal projection onto $\overline{A(X)}$ (which does indeed exist because $\overline{A(X)}$ is a closed subspace of $Y$). As you showed, $\overline {A(X)}\subset {A(X)^{\perp}}^{\perp}$ hence $P\varphi\in{A(X)^{\perp}}^{\perp}.$ Now ${A(X)^{\perp}}^{\perp}$ is a linear space (as is the orthogonal complement of any set) and hence $(\varphi-P\varphi)\in{A(X)^{\perp}}^{\perp}$. But the orthogonal projection $P$ is such that $\varphi=P\varphi+(\varphi - P\varphi)$ where $P\varphi\in\overline{A(X)}$ and $ (\varphi - P\varphi)\in{\overline{A(X)}^{\perp}}\subset {A(X)}^{\perp}$. But now since $$(\varphi - P\varphi)\in {A(X)^{\perp}}^{\perp}\cap {A(X)}^{\perp}=\{0\},$$ it follows that $\varphi=P\varphi\in\overline{A(X)}$, and the proof is finished in the way you demonstrated.
I think the most important place to improve your proof is here: "this means that $\varphi-P\varphi\perp A(X)$ and $\varphi-P\varphi\perp A(X)^{\perp},$ thus $\varphi-P\varphi$ must be in the closure." Instead you can immediately conclude there that  $(\varphi-P\varphi)=0.$ Indeed  we have that $(\varphi-P\varphi)\in {A(X)}^\perp$ and $(\varphi-P\varphi)\in {{A(X)}^\perp}^{\perp}$ so $$||\varphi-P\varphi||^2=\langle \varphi-P\varphi,\varphi-P\varphi\rangle=0.$$
