Inductor and Capacitor in Parallel
My book says the current through the inductor would pick up to 0.2A, while the current through the capacitor drops to 0A.
This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit.
Why? In DC steady state (the solution as $t\rightarrow\infty$), all the circuit voltages and currents are constant.
Now, recall that the voltage across an (ideal) inductor is given by
$$v_L = L\frac{di_L}{dt}$$
and so, since the inductor current is constant, the voltage across the inductor is zero. This is why you can replace the inductor with a wire.
For the (ideal) capacitor, the current through is given by
$$i_C = C \frac{dv_C}{dt}$$
and so, since the capacitor voltage is constant, the current through the capacitor is zero. This why you can replace the capacitor with an open-circuit.
In this case, it follows that both the capacitor current and voltage are zero in DC steady state.
Is my book wrong in saying after an infinite amount of time with the switch closed there will be a 10-volt difference across the capacitor?
Yes, if your book states that the capacitor has non-zero voltage across at infinite time, it is wrong for the reason I give above.
Tangential addendum:
From what I understand, the capacitor will charge up to 10V almost instantaneously, while no current will flow through the inductor.
That's not correct. As the capacitor charges, the current through the inductor must increase and this inductor current means that the capacitor voltage can never reach 10V (that would require zero inductor current). This could shown by solving for the step response of the capacitor voltage which is beyond the scope of the question. However, this circuit is easily simulated with LT Spice and I've attached a plot of the capacitor voltage just after the switch is closed. See that the maximum voltage is not quite 4V.
I think your book is correct. After an infinite amount of time with the switch closed there will be no potential difference across the capacitor, there will be no current through the capacitor and there will be stationary current $0.2$ A through the inductor and the resistor. Do you see any contradictions in this picture?
Good question, as the cap is charging the inductor back emf is decreasing, back emf is 0 at infinity time. Yes the cap will certainly have 10v as long as the switch is closed. The interesting part is what happens to the charge on the cap, I believe it eventually goes to zero because the bottom of the cap will be at 10v at an infinite time.