Let $H$ and $K$ be Hilbert spaces and $H \otimes K$ their Hilbert space tensor product.
Given $S \in B(H)$ and $T \in B(K)$, there is a unique operator $S \otimes T \in B(H \otimes K)$ such that $$(S \otimes T) (h \otimes k) = Sh \otimes Tk, \quad h\in H, k \in K.$$
Is the following equality true?
$$B(H \otimes K) = \overline{\text{span}}\{S \otimes T: S \in B(H), T \in B(K)\}$$
If so, how can one show this?
You need to be careful when you write $B(H)\otimes B(K)$ because, as a C$^*$-algebra, $B(H)$ is not nuclear and so there not a unique tensor norm. The product also makes sense as von Neumann algebras, in which case it is unique; the latter is the most common interpretation (but then the set in your title and the set in your question are not the same).
The equality is true if you take the sot-closure. In that case the closed span is $B(H)\otimes B(K)$ as von Neumann algebras, and you have $$(B(H)\otimes B(K))'=B(H)'\otimes B(K)'=\mathbb C\otimes \mathbb C,$$ so $B(H)\otimes B(K)=(B(H)\otimes B(K))''=(\mathbb C\otimes \mathbb C)'=B(H\otimes K)$. Alternatively, one can show that rank-one operators in $B(H\otimes K)$ are in $B(H)\odot B(K)$, and so the sot-closure will be everything.
With norm closure, on the other hand, there is no equality. This can be easily seen when both $H,K$ are separable, from the fact that the C$^*$-algebra $\overline{\operatorname{span}} B(H)\odot B(K)$ has several ideals ($K(H)\otimes B(K)$, $B(H)\otimes K(K)$, $K(H)\otimes K(K)$) while $B(H\otimes K)$ has only one.
Edit: When $H$ is not separable the idea in the last paragraph still works, by noting that the ideals of $B(H\otimes K)$ are generated by the operators with rank at most a certain cardinality, while either $K(H)\otimes B(K)$ or $B(H)\otimes K(K)$ will have operators with rank of full cardinality.