Let $A:L^2([0,1])\to L^2([0,1])$ given by $$ Af(t)=\int_0^1K(s,t)f(s)ds, $$ where $K$ is a mensurable square integrable operator, i.e $\int_0^1\int_0^1|K(s,t)|^2\,dsdt<\infty$.
$A$ is acompact operator: It is known that this is compact operator. In fact, we consider first the case where $K$ is a contínuous Kernel, and observe that $A$ maps $L^2([0,1])$ into $C^0([0,1])$, and the we use the Arzelá Ascoli Theorem in order to conclude tha $A$ is compact. In the general let $K$ a mensurable square integrable Kernel, consider a sequênce $(K_n)$ of contínuos kernels such that $K_n\to K$ in $L^2([0,1]^2)$ norm, this implies that $A_n\to A$ in the operator norm, where $$ A_nf(t)=\int_0^1K_n(s,t)f(s)ds, $$
and then we use que the space of the compact operators is closed.
My question ($A$ is a simetric operator?): My goal is to show that the operator $ A $ is symmetric , in order to show its worth the spectral theorem for $ A $. Please would you criticize my "proof": using the Fubini theorem we have for all $f,g\in L^2([0,1])$ that
\begin{eqnarray} \left<Af,g\right>&=&\int_{0}^{1}Af(t)g(t)dt=\int_{0}^{1}(\int_{0}^{1}K(s,t)f(s)ds)g(t)dt \\ & = & \int_0^1(\int_0^1K(s,t)g(t)dt)f(s)ds \\ & = & \left<f,Ag\right> \end{eqnarray} therefore $A=A^{*}$. My doubt in this demonstration consists in the fact that I assert that $A(s)=\int K(s,t)g(t)dt$, i.e I changed the variable of integration in the definition of A, is that correct??
No; note the definition of $A$ is $\int_0^1 K(s,t)f(s)\,ds$, and what you have written in the next to last line is $\int_0^1 K(s,t)g(t)\,dt$ (as noted by Daniel Fischer above), which is, in fact, just the definition of $A^*g$.
For a counterexample let $K(t,s) = t$ and $f \equiv 1$, $g = t$. Then
$$Af = \int_0^1 t \,ds = t$$
and so
$$\langle Af,g\rangle = \langle t, t\rangle = \int_0^1 t^2 \,dt = \frac13.$$
On the other hand,
$$Ag = \int_0^1 ts \,ds = \frac12 t,$$
so
$$\langle f,Ag\rangle = \langle 1, \frac12 t\rangle = \int_0^1 \frac12 t \,dt = \frac14.$$