Let $H$ be a complex Hilbert space and $\mathcal{L}(H)$ be the algebra of all bounded linear operators on $E$.



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> If $A,B\in \mathcal{L}(H)$, It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$?
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I try as follows:


Let $z\in \overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}$, then $z=\displaystyle\sum\_{i=1}^dx\_i\otimes y\_i$, where $x\_i\in\overline{\text{Im}(A)},\;y\_i\in\overline{\text{Im}(B)}$ and $d\in\mathbb{N}$. So, there exists $(\alpha\_i(n))\_{n\in\mathbb{N}}\subset \text{Im}(A)$ and $(\beta\_i(n))\_{n\in\mathbb{N}}\subset \text{Im}(B)$ such that $x\_i=\lim\_{n\rightarrow\infty}\alpha\_i(n)$ and $y\_i=\displaystyle\lim\_{n\rightarrow\infty}\beta\_i(n)$. Moreover, $\alpha\_i(n)=A\tilde{\alpha}\_i(n),\;\beta\_i(n)=B\tilde{\beta}\_i(n)$ and
$$\alpha\_i(n)\otimes\beta\_i(n)=(A\otimes B)(\tilde{\alpha}\_i(n)\otimes \tilde{\beta}\_i(n))\subset \text{Im}(A\otimes B),$$
where $(\tilde{\alpha}\_i(n))\_{n\in\mathbb{N}}\subset \mathcal{H}$ and $(\tilde{\beta}\_i(n))\_{n\in\mathbb{N}}\subset \mathcal{H}$. Let
$$z\_n=\displaystyle\sum\_{i=1}^d\alpha\_i(n)\otimes \beta\_i(n)\in \text{Im}(A\otimes B).$$
Since $\|x\otimes y\|=\|x\|\|y\|$, then the map $(x,y)\longmapsto x\otimes y$ is continuous. So $z\_n\longrightarrow z$ as $n\longrightarrow \infty$, this implies that $z\in \overline{\text{Im}(A\otimes B)}$. As a consequence, $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$.



It is true that $\overline{\text{Im}(A)}\otimes \overline{\text{Im}(B)}\subset \overline{\text{Im}(A\otimes B)}$?
The proof is okay. You did not specify the norm on the tensor product $H⊗H$
in which you take the closure. I do not know which one you want. Most naturally this would be the Hilbert-space or $\ell^2$ tensor product (which also describes the space of Hilbert-Schmidt operators). But the proof works for all reasonable norms. Another one describes the space of compact operators. Some of these are called Schatten-norms. See the beginnig of [here](http://www.mat.univie.ac.at/~michor/banach_modules_and_functors.pdf), or also section 7 of [here](http://www.mat.univie.ac.at/~michor/bookLN.pdf) (and references therein) for more information.


Added:
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Details of the proof:


The Hilbert-Schmidt norm on $\mathcal H\otimes\mathcal H$ can be defined as:
$$
\|z\|\_2 = \inf \Big\{ \|(\|x\_i\|)\|\_{\ell^2}.\sup\_{\|y'\|\le1}\|(\langle y\_i,y'\rangle)\|\_{\ell^2} : z = \sum x\_i\otimes y\_i \Big\}
$$
Here the infimum is taken over all (finite) representations of $z$ in $\mathcal H\otimes\mathcal H$. This can be seen from section 7 of the second reference I gave above (quite complicated there); best are the references Chevet or Saphar therein which I do not have available now. 
Given $z$ as in your proof above, choose a finite representation $z = \sum\_{i=1}^{N(n)} x\_i(n)\otimes y\_i(n)$ such that 
$$
\|(\|x\_i(n)\|)\|\_{\ell^2}.\sup\_{\|y'\|\le1}\|(\langle y\_i(n),y'\rangle)\|\_{\ell^2}
\le \|z\|\_2 +\frac{\epsilon}n\,.
$$
Now choose your $\alpha\_i(n,m)$ and $\beta\_i(n,m)$ in such a way that 
$$
\|(\|x\_i(n)-\alpha\_i(n,m)\|)\|\_{\ell^2}<\frac{\epsilon}{nm}\,,\qquad
\|(\|y\_i(n)-\beta\_i(n,m)\|)\|\_{\ell^2}<\frac{\epsilon}{nm}\,.
$$
Then you can prove convergence in the $\ell^2$-tensor product.


