 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 $\epsilon >0$, choose $n$ with $\Vert T-T_n\Vert <\epsilon /3$ and a finite $\epsilon /3$-net for $T_n$ of the unit ball. The same centres give an $\epsilon$-net for $T$ of that ball. Its closure is totally bounded and complete in the Banach space $Y$, hence compact. Thus 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 $\mathrm{span}\{f_i\}$ and is bounded. MoreoverConsequently $T_n\to T$ in operator norm, proving that the stated integral operator is compact.