I feel like this should be true. Let $\mathrm{range}(E)=A$ and $u$ be an arbitrary vector in a Hilbert space $H$, it is sufficient to show $\langle (E-\overline{E})u,u\rangle=0$. By Cauchy-Schwarz:
$$\langle (E-\overline{E})u,u\rangle\le ||Eu-\overline{E}u||\cdot||u||$$
$\overline{E}u$ is either in $A$ or a limit point of $A$, in the former case $Eu=\overline{E}u$ and we are done. For the latter though I am having some trouble. I feel like we should have $Eu=\overline{E}u$. To show this my idea is to decompose $$H=\overline{A}\oplus\overline{A}^\perp$$
Then we can write
$$u=u_0+u_1$$ where $u_0\in\overline{A}$ and $u_1\in\overline{A}^\perp$. We can then give $\overline{A}$ the same Schauder basis as $A$ (as the basis for $A$ is dense in $\overline{A}$), i.e. $$u_0=\sum_i c_i a_i$$ for constants $c_i$ and $\{a_i\}$ an orthonormal basis of $A$. Then $Eu=u_0=\overline{E}u$.
This however brings up the problem of defining the projection onto a non-closed subspace, it seems like the range of a projection operator must be closed and so $E$ has range $\overline{A}$ rather than just $A$. Is this true?
An operator $P$ on a Hilbert space is an orthogonal projection iff $$ P^{2}=P=P^{\star}. $$ Such a projection always has a closed range because $\mathcal{R}(P)=\mathcal{N}(I-P)$.
You can try to define the orthogonal projection $P_{\mathcal{M}}x$ of a vector $x$ onto a subspace $\mathcal{M}$ as the unique $m \in \mathcal{M}$ such that $(x-m)\perp \mathcal{M}$. However, such a thing is not defined for $x \in \overline{\mathcal{M}}\setminus\mathcal{M}$ because an orthogonal projection $m \in \mathcal{M}$ would be such that $(x-m)\perp\mathcal{M}$. In particular $(x-m)\perp m$. But also, because $x \in \overline{\mathcal{M}}$, then $(x-m)\perp x$. So $(x-m)\perp(x-m)$, which implies that $x=m$, which is a contradiction because $x \notin\mathcal{M}$.