.  $Y \cap (S \times \{n\}) = \{x_1, x_2\} \times \{n\}$.

In Case 1, each $\mathbb{Z}_{k+1}^{2}$ layer contains exactly one point of $Y$. $T$ then tiles the rest of the layer by Proposition \[onepoint\].

In Case 2, some of the layers contain two points of $Y$, and some of the layers contain no points. Holes of size 0 and 2 do not exist, so we will need copies of $T$ in the third direction to fill some gaps (where $Y$ consists of copies of $T$ in the fourth direction). The following lemma provides us with a way to do this:

\[otherlemma\] Let $A \subset \mathbb{Z}^d$, $|S| = 3k$. Then there exists $B \subset S \times \mathbb{Z}$ such that $T$ tiles $B$, and $$|B \cap (S \times \{n\})| = 
    \begin{cases}
    k+1 & \text{\emph{if} } n \equiv 1, \ldots, k \pmod{2k}\\
    k-1 & \text{\emph{if} } n \equiv k+1, \ldots, 2k \pmod{2k}
    \end{cases}$$

Let $A = \{a_1, \ldots, a_{3k}\}$. Then:\
For $i = 1, \ldots, k$, place a copy of $T$ beginning at $\{a_i\} \times \{n\}$ for every $n \equiv i \pmod{6k}$.\
For $i = k+1, \ldots, 2k$, place a copy of $T$ beginning at $\{a_i\} \times \{n\}$ for every $n \equiv i+k \pmod{6k}$.\
For $i = 2k+1, \ldots, 3k$, place a copy of $T$ beginning at $\{a_i\} \times \{n\}$ for every $n \equiv i+2k \pmod{6k}$.\
We now observe that the union $B$ of these tiles has the required property.\
For $n \equiv 1, \ldots, k \pmod{6k}$, $B \cap (A \times \{n\}) = \{a_{2k+n}, \ldots, a_{3k}, a_1, \ldots, a_n\}$ (size $k+1$).\
For $n \equiv k+1, \ldots, 2k \pmod{6k}$, $B \cap (A \times \{n\}) = \{a_1, \ldots, a_k\}\setminus\{a_{n-k}\}$ (size $k-1$).\
For $n \equiv 2k+1, \ldots, 3k \pmod{6k}$, $B \cap (A \times \{n\}) = \{a_{n-2k}, \ldots, a_{n-k}\}$ (size $k+1$).\
For $n \equiv 3k+1, \ldots, 4k \pmod{6k}$, $B \cap (A \times \{n\}) = \{a_{k+1}, \ldots, a_{2k}\}\setminus\{a_{n-2k}\}$ (size $k-1$).\
For $n \equiv 4k+1, \ldots, 5k \pmod{6k}$, $B \cap (A \times \{n\}) = \{a_{n-3k}, \ldots, a_{n-2k}\}$ (size $k+1$).\
For $n \equiv 5k+1, \ldots, 6k \pmod{6k}$, $B \cap (A \times \{n\}) = \{a_{2k+1}, \ldots,