= Solution
The real <Stone-Weierstrass theorem> says that a subalgebra of $C(X,\mathbb R)$ on a compact metric space is uniformly dense if it contains the constants and separates points. The finite sums $\sum_i f_i(x)g_i(y)$ form such an algebra on $[0,1]^2$: multiplication preserves this form, the constant one belongs to it, and the coordinate functions distinguish every pair of different points. Thus they approximate the continuous kernel uniformly.
A <finite-rank operator> has finite-dimensional range. A <compact operator> maps the unit ball to a set with compact closure, equivalently sends every bounded sequence to a sequence with a convergent subsequence. The identity on $C[0,1]$ is bounded but not compact: the functions $x^n$ have supremum norm one, and every subsequence has the discontinuous pointwise limit which is zero on $[0,1)$ and one at one. It therefore has no uniformly convergent subsequence.
A bounded finite-rank operator is compact because bounded sets in its finite-dimensional range have compact closure. Suppose $T_n\to T$ in <operator norm>. For any $\varepsilon>0$, choose $n$ with $\|T-T_n\|<\varepsilon/3$ and a finite $\varepsilon/3$-net for $T_n$ of the unit ball. The same centres give an $\varepsilon$-net for $T$ of that ball. Its closure is totally bounded and complete in the <Banach space> $Y$, hence compact. Thus \b[an operator-norm limit of compact operators is compact].
For $K_n(x,y)=\sum_i f_i(x)g_i(y)$ uniformly approximating $K$, the integral operator $T_n$ has range in $\operatorname{span}\{f_i\}$ and is bounded. Moreover
$$
\|(T-T_n)f\|_\infty\leq\|K-K_n\|_\infty\|f\|_\infty.
$$
Consequently $T_n\to T$ in operator norm, proving that \b[the stated integral operator is compact].
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