It may be worth noting the phenomena that can appear in Hilbert spaces, where study of the things is more decisive, both positive and negative.

First, I like the "definition" of "trace class" $T:X\rightarrow Y$ with Hilbert spaces $X,Y$ to be that $T$ is a composition of two Hilbert-Schmidt operators (which are defined as being in the HS-norm completion of the algebraic tensor product $X^*\otimes_{\mathrm {alg}}Y$. This gives an intrinsic definition... which, if desired, is provably equivalent to the (ugly) requirement that $\sum |\langle Tx_i,y_i\rangle| <\infty$ for every pair of orthonormal bases.

The reason I recall this cliche is that, in many applications of interest (to me!), natural operators are visibly Hilbert-Schmidt (if compact at all), and the issue becomes to prove trace-class. In practice (for me) it often happens that we know that every one of these integral operators *is* a finite sum of compositions of two such, proving trace-class.

Sometimes proof of the latter is highly non-trivial, as in the Cartier/Dixmier-Malliavin proof that test functions on Lie groups are finite linear combinations of convolutions of pairs of such. The totally-disconnected group analogue is trivial.

That summing or integrating down the diagonal fails is easy to illustrate with not-normal operators: the shift operator on one-sided or two-sided $\ell^2$ might seem to have trace absolutely summing to $0$, but it is not trace class at all. Integral analogues of this are clear.

Edit: in response to question about reference, etc.: in Lang's "SL(2,R)" the equivalence of the coordinate-dependent definition of "trace class", and the definition as composition of two Hilbert-Schmidt, are carefully compared. Further, in that same source, various conditions on a kernel assuring that its trace is equal to its integral over the diagonal are carefully treated. (I must say "... in contrast to dangerously glib treatments elsewhere").

Further edit: in response to Yemon Choi's comments: yes, the space of trace-class operators is also the closure of finite-rank operators with respect to the "trace norm"... At the moment, verification of the equivalence seems straightforward.