The infinite grid of unit resistors has an equivalent resistance of $\frac12$ between adjacent nodes. The rules for addition of capacitors in series and parallel are the same for addition of resistors in parallel and series respectively. So we expect the equivalent capacitance of an infinite grid of unit capacitors to be the inverse of the result for resistors, ie 2.
The same argument works for equivalent capacitance between any two nodes $(0,0)$ and $(m,n)$. If the equivalent resistor network has resistance $R_{m,n}$ units then the capacitor network has capacitance $C_{m,n}=1/R_{m,n}$ units between the same nodes.
The trick for the infinite resistor network is to use symmetry and superposition.
Imagine case A : that current $I_{in}=1A$ enters the network at node A, spreads out symmetrically in every outward direction and comes out at a circular conducting ring at infinity. Separately there is case B : current $I_{out}=1A$ enters at the conducting ring at infinity, spreads inward symmetrically through the network and comes out at node B. Case B is the reverse of case A, shifted along by one node, so the currents in corresponding branches are the same.
In case A, $1A$ is injected at A so by symmetry the current in branch AB will be $\frac14A$. In case B we have the same result that a current of $\frac14A$ flows along branch AB in the same direction AB.
The problem we have is the superposition of cases A and B : a current of $1A$ flows in at A and exits at B. The currents in and out at infinity ($\frac34 A$) cancel out so that all current which goes in at A comes out at B.
In branch AB the two currents are in the same direction, making a total of $\frac12 A$. If branch AB has unit resistance then the voltage across the network between AB is $V_{out}=\frac12V$. The equivalent resistance of the network between A and B is $$R_{AB}=\frac{V_{out}}{I_{in}}=\frac{\frac12 V}{1A}=\frac12 \Omega$$
The same trick works for capacitance, except that this time we are injecting charge instead of current.
We now inject charge of $Q_{in}=+1\mu C$ at A (case A). Without any charge removed at B the charges on each capacitor adjacent to A, especially that in branch AB, are $\pm \frac14 \mu C$.
In case B we remove charge $Q_{out}=+1\mu C$ at B, which is equivalent to injecting charge $-1\mu C$. The charges on each capacitor adjacent to B, including that along AB, are $\pm \frac14 \mu C$, again with the +ve side connected to A and the -ve side to B.
Superposing the two cases (injection at A, removal at B) the total charge on capacitor AB is $Q_{AB}=\frac12 \mu C$ and the voltage across this unit capacitor is $V_{out}=\frac12 V$. The equivalent capacitance of the network is $$C_{eq}=\frac{Q_{in}}{V_{out}}=\frac{1\mu C}{\frac12 V}=2 \mu F$$
