Compact operators in $\bigoplus_{i \in I} H_i$ Let $\{H_i: i \in I\}$ be a collection of Hilbert spaces. We can form the Hilbert space direct sum
$$H:= \bigoplus_{i \in I} H_i$$
Question: Is there a "nice" dense subset of $B_0(H)$ (compact operators on the direct sum $H$)? For example, something like the operators
$$\left\{\bigoplus_{i \in I}  T_i \ \Bigg| \ T_i \in F(H_i), \text{all but  finitely many $T_i$ are zero}\right\}$$
where $F(H_i)$ are the finite-rank operators on $H_i$.
 A: Let me begin with a discussion of "operators vs. matrices", striving to make precise statements based on adjoints and
compositions   of operators, while avoiding the need to make  identifications between  sets, such as between a factor of a direct sum and its
canonical image.

Let $\{H_i\}_{ i \in I}$ be a collection of Hilbert spaces and put
$$
  H:= \bigoplus_{i \in I} H_i.
  $$
For each $i$, denote by $\iota _i:H_i\to H$ the  inclusion map and   observe that $\iota _i^*$ is precisely the projection from
$H$ onto $H_i$.
Incidentally, since $\iota _i$ is an isometry, then $\iota _i^*\iota _i$ is the identity on the domain of $\iota _i$ , while $\iota _i\iota _i^*$ is the orthogonal
projection onto $\iota_i (H_i)$, a.k.a. the canonical image of $H_i$ in $H$.  See $(\dagger)$ below regarding
a subtle conceptual distinction between $\iota _i\iota _i^*$ and $\iota _i^*$.
Given any bounded linear operator $T\in B(H)$, put
$$
  T_{i,j}=\iota _i^*T\iota _j, \quad \forall i,j\in  I,
  $$
so that each $T_{i,j}\in  B(H_j,H_i)$.  The "matrix" $(T_{i,j})_{i,j}$ may then be seen as the matrix of $T$ relative to the
above decomposition of $H$, and it can be used to recover $T$ in the following way:
given any vector $\xi \in  H$, write $\xi =(\xi _i)_{i\in I}$, so that each $\xi _i=\iota _i^*\xi $.
Noticing that $\xi =\sum_{j\in I} \iota _j\xi _j$, one then has for every $i$   that
$$
  \iota _i^*T\xi  = \sum_{j\in I} \iota _i^*T\iota _j\xi _j,
  $$
so, writing $T(\xi )=(\eta _i)_{i\in I}$, we see that the $i^{th}$ component $\eta _i$ of $T(\xi )$  may be computed in the familiar way as
$$
  \eta _i=\sum_{j\in I} T_{i, j}\xi _j.
  $$
Conversely, if $(T_{i,j})_{i,j}$ is any given matrix, with $T_{i,j}\in  B(H_j,H_i)$,  assume that $(T_{i,j})_{i,j}$ has finitely many
nonzero entries.   One can then define a bounded operator $T$ on $H$ by
$$
  T = \sum_{i, j} \iota _iT_{i,j}\iota^* _j,
  \tag 1
  $$
and it is easy to see that the matrix of $T$ coincides with the originally given matrix.
In case one moreover has that each $T_{i,j}$ lies in the set $K(H_j,H_i)$ of compact operators,  then $T$ is clearly
compact.  If instead each $T_{i,j}$ lies in the set $F(H_j,H_i)$ of finite rank operators, then $T$ has finite rank.

Directly addressing the question posed by the OP, the set formed by all operators built as in (1), from finitely
supported matrices $(T_{i,j})_{i,j}$, with each $T_{i,j}\in F(H_j,H_i)$, is a dense subspace of $K(H)$.  The proof may be
easily done by observing that the identity operator $1_H$ may be writen as
$$
  1_H = \sum_i \iota _i\iota _i^*,
  $$
with convergence in the strong operator topology,   and moreover noticing that left- or right-multiplication by compact operators turns
strong convergence into norm convergence.

$(\dagger)$ There is a subtle conceptual distinction between $\iota _i\iota _i^*$ and $\iota _i^*$ in the sense that both project
onto $H_i$, but the former  has $H$ for co-domain while the latter's  co-domain is $H_i$.  Co-domains are often
irrelevant in Math, but since the co-domain of an operator  becomes the domain of its adjoint, it is sometimes important to pay
attention to co-domains too!  In this note  all compositions $TS$ of operators are such that the domain of $T$ is equal
to the co-domain of $S$, even though for $TS$ to make sense, it is enough for the domain of $T$ to contain the range of $S$.
