Why does the copper anode dissolve?
The anode is where oxidations happen (the cathode is where reduction takes place).

In the vicinity of your anode, you have the following compounds: $\ce{H2O}$, $\ce{Cu}$, $\ce{Cu^2+}$, $\ce{SO4^2-}$. If you check oxidation states, you will realise that oxygen is always $\mathrm{-II}$, hydrogen is $\mathrm{+I}$, sulphur is $\mathrm{+VI}$ and copper is present in both $\mathrm{\pm 0}$ and $\mathrm{+II}$. Thus, the only species available for oxidation are copper($0$) and oxygen($\mathrm{-II}$).

Checking standard potentials:
\begin{align}
\ce{O2 + 4 H+ + 4 e- &amp;&lt;=&gt; 2 H2O} \qquad &amp;E^\circ = \pu{+1.23 V}\tag{1}\\
\ce{Cu^2+ + 2 e- &amp;&lt;=&gt; Cu} \qquad &amp;E^\circ = \pu{+0.35 V}\tag{2}
\end{align}

It should be clear why copper is oxidised.

On the reduction side, hydrogen, copper and sulphate all could accept electrons. Copper is again the happiest of the three to accept electrons.