Here's a reformulation of [my other answer](https://mathoverflow.net/a/420104) in terms of compactness and general topological abstract nonsense, for those who are interested in the general template for this style of argument:

For a given finite polyomino which can cover squares of arbitrary size, consider the set $S$ of all possible ways to place this polyomino onto an infinite grid. A subset $Q$ of $S$ can be considered a *placement* of copies of the polyomino onto the plane, where gaps or overlap may occur. A placement $Q$ may be represented by its characteristic function $\chi_Q: S \to \{0,1\}$.

The space $P := \mathcal P(S)$ of all possible placements has a canonical bijection to the space $\{0,1\}^S$ of all characteristic functions. The latter set is a cartesian product of two-element sets indexed by $S$, so we can equip it with the topology of a direct product of discrete two-point spaces indexed by  $S$. Two-point spaces are compact, so the resulting topology on $\mathcal P(S)$ is also compact.

The topology of this direct product is generated by the clopen sets $A_s := \{Q \subset S: s \in Q\}$ for $s \in S$ and their complements $A_s^c$.

For any cell $x$ of the grid, we can define the subset $P_x$ of placements which are *valid* on $x$, meaning that exactly one polyomino of the placement covers the cell $x$. Observe that since the set $S_x := \{s \in S: x \in s\}$ is finite, we can describe

$$P_x = \bigcup_{s \in S_x}\left(A_s \cap \bigcap_{t \in S_x\setminus\{s\}}A_t^c\right)$$

as a finite union of intersections of closed sets, so $P_x$ is a closed subset of $P$.

Now let $B_1 \subset B_2 \subset \ldots$ be a sequence of larger and larger squares of grid cells whose union is the entire plane. For each $B_i$ we can define the set $Q_i := \bigcap_{x \in B_i} P_x$ of placements valid within the square $B_i$. As an intersection of closed subsets of $P$, the $Q_i$ are themselves closed subsets of $P$, and since $P$ is compact so are the $Q_i$. Observe that $B_i \subset B_j$ implies $Q_i \supset Q_j$, and note that each $Q_i$ nonempty because of our initial assumption that the polyomino we chose can cover every square $B_i$ somehow.

Now the $Q_i$ are an infinite descending nested sequence of nonempty compact sets. This lets us apply [Cantor's intersection theorem](https://en.wikipedia.org/wiki/Cantor%27s_intersection_theorem), the core of most compactness arguments out there, which simply tells us that the intersection of all $Q_i$ is nonempty.

However, the intersection of all $Q_i$ is also the intersection of all $P_x$, and if we recall how we defined those $P_x$, namely that $P_x$ only contains placements valid on the cell $x$, we notice that we just proved the existence of a placement that is valid on every single cell of the plane, also known as a tiling of the plane by copies of the polyomino. $~~\square$

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Of note here is that it is essential for this proof that the polyomino in question is *finite*, otherwise the $P_x$ are not necessarily closed. In fact, there are many classes of *infinite* polyominos which can tile arbitrary squares but not the entire plane. The simplest of these infinite polyominos is probably just a whole plane with a single cell missing, but there are many pretty counterexamples. Finding some is left as an exercise to the reader.