The argument is sloppy since it allows $\Gamma$ to intersect $K$. A clean way to argue is as follows.

1. Define $\eta$ as in Rudin's paper (i.e. every point of $K$ is at distance $\ge 2\eta$ from the complement to $\Omega$) and $\epsilon=\eta/3$. Let $K_\epsilon$ denote the closed $\epsilon$-neighborhood of $K$, i.e. the set of points $x$ such that the minimal distance from $x$ to $K$ is $\le \epsilon$. Then $K_\epsilon\subset \Omega$ ($\Omega=U$ in your notation) and $K$ is contained in the interior of $K_\epsilon$. Then argue almost as in Rudin and consider those squares with edge-size $\epsilon$ which intersect $K_\epsilon$. The union $W$ of these squares is still contained in $\Omega$ and $K$ is contained in the interior of $W$. To see that $W\subset Q$ note that every point of $W$ is within distance at most $2\sqrt{2}\epsilon$ from a point in $K$. But $2\sqrt{2}\epsilon< 2\eta$ since $\epsilon=\eta/3$.

2. Then form the 2-dimensional chain $$ C=\sum_{j} Q_j $$ summing up all the squares used above with the orientation induced from the complex plane. Consider the (algebraic) boundary $\partial C$ of this chain: $$ \partial C= \sum_j \partial Q_j $$ where each $\partial Q_j$ is what you denoted $\delta_j$. This is your 1-cycle $\Gamma$. The key fact is that $$ \partial (\partial c)=0 $$ for each 2-chain. Since $\partial$ is a linear operator, to verify this identity you just check it for each square $Q$, where it is manifestly true: $$ c= \sum_j a_j Q_j, a_j\in \mathbb Z, $$ $$ \partial c= \sum_j a_j \partial Q_j, $$ $$ \partial \partial c= \sum_j a_j \partial \partial Q_j=0. $$ (This is a baby version of what you see in an algebraic topology class if you ever take one.) Thus, $\partial \Gamma=0$. The fact that $\Gamma$ is a cycle in $\Omega$ follows from the fact that $W\subset \Omega$ and, hence, every edge of a square in $W$ is also contained in $\Omega$. The fact that $\Gamma$ is disjoint from $K$ follows from the fact that $\Gamma$ is an algebraic sum of edges each of which is contained in the topological boundary of $W$ and the latter is disjoint from $K$ by the construction.

The argument is sloppy since it allows $\Gamma$ to intersect $K$. A clean way to argue is as follows.

1. Define $\eta$ as in Rudin's paper (i.e. every point of $K$ is at distance $\ge 2\eta$ from the complement to $\Omega$) and $\epsilon=\eta/3$. Let $K_\epsilon$ denote the closed $\epsilon$-neighborhood of $K$, i.e. the set of points $x$ such that the minimal distance from $x$ to $K$ is $\le \epsilon$. Then $K_\epsilon\subset \Omega$ ($\Omega=U$ in your notation) and $K$ is contained in the interior of $K_\epsilon$. Then argue almost as in Rudin and consider those squares with edge-size $\epsilon$ which intersect $K_\epsilon$. The union $W$ of these squares is still contained in $\Omega$ and $K$ is contained in the interior of $W$. To see that $W\subset \Omega$ note that every point of $W$ is within distance at most $2\sqrt{2}\epsilon$ from a point in $K$. But $2\sqrt{2}\epsilon< 2\eta$ since $\epsilon=\eta/3$.

2. Then form the 2-dimensional chain $$ C=\sum_{j} Q_j $$ summing up all the squares used above with the orientation induced from the complex plane. Consider the (algebraic) boundary $\partial C$ of this chain: $$ \partial C= \sum_j \partial Q_j $$ where each $\partial Q_j$ is what you denoted $\delta_j$. This is your 1-cycle $\Gamma$. The key fact is that $$ \partial (\partial c)=0 $$ for each 2-chain. Since $\partial$ is a linear operator, to verify this identity you just check it for each square $Q$, where it is manifestly true: $$ c= \sum_j a_j Q_j, a_j\in \mathbb Z, $$ $$ \partial c= \sum_j a_j \partial Q_j, $$ $$ \partial \partial c= \sum_j a_j \partial \partial Q_j=0. $$ (This is a baby version of what you see in an algebraic topology class if you ever take one.) Thus, $\partial \Gamma=0$. The fact that $\Gamma$ is a cycle in $\Omega$ follows from the fact that $W\subset \Omega$ and, hence, every edge of a square in $W$ is also contained in $\Omega$. The fact that $\Gamma$ is disjoint from $K$ follows from the fact that $\Gamma$ is an algebraic sum of edges each of which is contained in the topological boundary of $W$ and the latter is disjoint from $K$ by the construction.