The definition of Hilbert-Schmidt operator should still be valid even when the Hilbert space is not separable: If $e_i$ for $i\in I$ is an orthonormal basis for a Hilbert space, and
$${\rm tr}(T^*T)=\sum_{i\in I}\|Te_{i}\|^{2}<\infty$$
Then $T$ is Hilbert-Schmidt. Of course if the sum is finite then there can only be countably many non-zero terms in the summation. However, I am not sure how to show that such operators are compact.
There are many ways to prove compactness, e.g., via density of the finite-rank operators or the singular value decomposition. A more basic one—which does not resort to the definition of compactness—is via the following characterization of compactness taken from "Introduction to Functional Analysis" by Meise & Vogt (1997):
Lemma 16.17. The following are equivalent for every self-adjoint operator $T \in B(H)$:
- $T$ is compact.
- For every countable orthonormal system $(e_j)_{j \in \mathbb{N}}$ in $H$ we have $ \lim_{j \to \infty} \langle e_j,T e_j\rangle = 0$.
(Be aware that the cited book does not assume separability of $H$)
Anyway, with this lemma in mind you can argue as follows: Starting from a Hilbert-Schmidt operator $T$ decompose it as $T=T_1+iT_2$ where $T_1=\frac12(T+T^*)$, $T_2=\frac1{2i}(T-T^*)$ are self-adjoint and, by linearity, Hilbert-Schmidt. Thus it suffices to prove that $T_1,T_2$ are both compact, because then the same is true for any linear combination. Now let $k=1,2$ and let $(e_j)_{j \in \mathbb{N}}$ be an arbitrary orthonormal system in $H$. Completing it to an orthonormal basis $(e_j)_{j\in J}$ of $H$ we find $$ \sum_{j\in\mathbb N}\|T_ke_j\|^2\leq\sum_{j\in J}\|T_ke_j\|^2={\rm tr}(T_k^*T_k)<\infty $$ meaning that, necessarily, $\lim_{j\to\infty}\|T_ke_j\|\to 0$. Cauchy-Schwarz now implies $$ \langle e_j,T_ke_j\rangle\leq|\langle e_j,T_ke_j\rangle|\leq\|e_j\|\|T_ke_j\|=\|T_ke_j\|\to 0\qquad\text{as }j\to\infty\,. $$ But $(e_j)_{j\in\mathbb N}$ was arbitrary so the above Lemma implies that $T_1,T_2$ are compact, hence so it $T$. $\square$
Preword
The Hilbert dimension plays no role at all!
Problem
A trace class operator is Hilbert Schmidt. (Decomposition)
A Hilbert Schmidt operator is compact. ([Denseness][2])
