I have been staring at this for several hours already, still unable to find my own error. I know, it is embarassing, but I need your help to spot it, please. Thank you.
Claim: Every bounded linear operator on $L^2$ has an integral kernel (and in fact is Hilbert-Schmidt).
Proof: Let $(X,m)$ be a space with measure, and let $U : L^2(X) \to L^2(X)$ be a bounded linear operator. Define $I : L^2(X) \otimes _{alg} L^2(X) \to \mathbb C$ (the tensor product being the algebraic one, not the topological one) by $$I(f \otimes g) = \int _X f \ Ug \ \mathrm d m \ .$$ Notice that $I$ is linear and that, using the Cauchy-Schwarz inequality, $$|I (f \otimes g)| \le \| f \| _{L^2} \ \| Ug \| _{L^2} \le \| U \| \ \| f \| _{L^2} \ \| g \| _{L^2} = \| U \| \ \| f \otimes g \| _{L^2(X \times X)} \ ,$$ which means that we may extend $I$ by continuity to the whole of $L^2(X \times X)$. By Riesz's theorem, it follows that there exists $k \in L^2 (X \times X)$ such that $I (F) = \int _{X \times X} \bar k \ F \ \mathrm d (m \times m)$. In particular, if $F = f \otimes g$, it follows that $$ \int _X f(x) \ (Ug)(x) \ \mathrm d m (x) = \int _X f(x) \left( \int _X \overline {k(x,y)} \ g(y) \ \mathrm d m (y) \right) \mathrm d m (x) \ ,$$ whence, since $f$ is arbitrary, it follows that $$ (Ug)(x) = \int _X \overline {k(x,y)} \ f(y) \ \mathrm d m (y) \ ,$$ for almost all $x$, which is obviously not true.
Where am I losing it?
The mistake, as shown by Olivier Bégassat in a comment, is hidden in the fact that I check the continuity of $I$ only on monomials of the form $f \otimes g$, not on sums of such monomials (as I should).