Notation
$H$ - some Hilbert space
$H^{cc}$ - complex conjugate Hilbert space, i.e. with multiplication on complex conjugate scalars
$H_1\otimes H_2$ Hilbert tensor product of Hilbert spaces $H_1$ and $H_2$
$\mathcal{F}(H)$ finite rank operators on $H$
$\mathcal{K}(H)$ compact operators on $H$
$\mathcal{B}(H)$ bounded operators on $H$
$x\bigcirc y$ rank one operator on $H$ well defined by $(x\bigcirc y)(z)=\langle z,y\rangle x$ where $x,y,z\in H$
$a\;\dot{\otimes }\;b$ Hilbert tensor product of operators $a\in\mathcal{B}(H_1)$ and $b\in\mathcal{B}(H_2)$ well defined by $(a\;\dot{\otimes }\;b)(x\otimes y)=a(x)\otimes b(y)$
Facts
$\mathcal{F}(H)=\operatorname{span}\{ x\bigcirc y:x\in H,\; y\in H\}$
$\mathcal{K}(H)=\operatorname{cl}_{\mathcal{B}(H)}\mathcal{F}(H)$
The proof given below is valid for all Hilbert spaces.
Since $\mathcal{K}(H)$ is a nuclear $C^*$ algebra for any Hilbert space $H$, then we can consider any $C^*$ norm on the algebraic tensor product $\mathcal{K}(H_1)\odot K(H_2)$. We will consider the spatial tensor norm, so $ \mathcal{K}(H_1)\otimes K(H_2)=\operatorname{cl}_{\mathcal{B(H_1\otimes H_2)}}(\operatorname{span}\{ a\;\dot{\otimes}\; b:a\in\mathcal{K}(H_1),\; b\in\mathcal{K}(H_2)\})\tag{1} $ where $a\;\dot{\otimes}\; b\in\mathcal{B}(H_1\otimes H_2)$ is well defined by equality $(a\;\dot{\otimes}\; b)(x\otimes y)=a(x)\otimes b(y)$. Denote the closed linear subspace in the right hand side of $(1)$ by $E$.
Lemma 1. $\mathcal{F}(H_1\otimes H_2)\subset E$.
Proof. Since bilinear operator $\bigcirc:H\times H^{cc}\to\mathcal{F}(H)$ is bounded, then $\mathcal{F}(H)=\operatorname{span}\{x\bigcirc y: x,y\in S\}$ for any $S\subset H$ such that $H=\operatorname{cl}_H(\operatorname{span}S)$. For $H=H_1\otimes H_2$ we can take $S=\{x\otimes y:x\in H_1,y\in H_2\}$. Now to prove that $\mathcal{F}(H_1\otimes H_2)\subset E$ it is remains to show that $(x\otimes y)\bigcirc (x'\otimes y')\in E$ for all $x,x'\in H_1$, $y,y'\in H_2$. But this is indeed true because $(x\otimes y)\bigcirc (x'\otimes y')=a'\;\dot{\otimes}\; b'$ for $a'=x\bigcirc x'\in\mathcal{K}(H_1)$ and $b'=y\bigcirc y'\in\mathcal{K}(H_2)$.
Lemma 2. $E\subset \mathcal{K}(H_1\otimes H_2)$.
Proof. Consider $a'=x\bigcirc x'\in\mathcal{F}(H_1)$ and $b'=y\bigcirc y'\in\mathcal{F}(H_2)$ for some $x,x'\in H_1$ and $y,y'\in H_2$. Recall $a'\;\dot{\otimes}\; b'=(x\otimes y)\bigcirc (x'\otimes y')\in\mathcal{F}(H_1\otimes H_2)$. Since $\dot{\otimes}$ is bilinear operator and $\mathcal{F}(H)=\operatorname{span}\{x\bigcirc y:x,y\in H\}$ for any Hilbert space $H$ then $a'\;\dot{\otimes}\; b'\in\mathcal{F}(H_1\otimes H_2)$. In other words $\dot{\otimes}\;(\mathcal{F}(H_1),\mathcal{F}(H_2))\subset \mathcal{F}(H_1\otimes H_2)$. Since the bilinear operator $\dot{\otimes}:\mathcal{B}(H_1)\times \mathcal{B}(H_2)\to\mathcal{B}(H_1\otimes H_2)$ is bounded, then $ \dot{\otimes}\;(\mathcal{K}(H_1),\mathcal{K}(H_2)) =\dot{\otimes}\;(\operatorname{cl}_{\mathcal{B}(H_1)}\mathcal{F}(H),\operatorname{cl}_{\mathcal{B}(H)}\mathcal{F}(H_2)) \subset\operatorname{cl}_{\mathcal{B}(H_1\otimes H_2)}\left(\dot{\otimes}\;(\mathcal{F}(H_1),\mathcal{F}(H_2))\right) \subset\operatorname{cl}_{\mathcal{B}(H_1\otimes H_2)}\mathcal{F}(H_1\otimes H_2) =\mathcal{K}(H_1\otimes H_2) $ So $E=\operatorname{cl}_{\mathcal{B}(H_1\otimes H_2)}\dot{\otimes}\;(\mathcal{K}(H_1),\mathcal{K}(H_2))\subset\mathcal{K}(H_1\otimes H_2)$.
Proposition. $E=\mathcal{K}(H_1\otimes H_2)$.
Proof. Since $\mathcal{K}(H)=\operatorname{cl}_{\mathcal{B}(H)}\mathcal{F}(H)$ for any Hilbert space $H$ and $E$ is cloed, then it is enough to show that $\mathcal{F}(H_1\otimes H_2)\subset E\subset\mathcal{K}(H_1\otimes H_2)$. Now the result follows from lemma $1$ and lemma $2$.