Horizontal bands with height exactly $1$, closed on the bottom, open on the top, provide such a coloring. Thus, $(x, y)$ is red if $2|\lfloor y \rfloor$ and blue otherwise.
Number the square's vertices in order of height and let $\theta$ be the angle between a horizontal line passing through vertex $1$ and the side of the square $\overline{12}$. Without loss of generality, $0 \leq \theta \leq \frac{\pi}{2}$. (Otherwise perform a reflection.) Let $y_1$ be the $y$-coordinate of vertex $1$. Then the $y$-coordinates of the four vertices are:
$$y_1\\
y_2 = y_1+\sin \theta\\
y_3=y_1+ \cos \theta \\
y_4=y_1+\sin \theta + \cos \theta = y_1+\sqrt{2}\sin (\theta + \frac{\pi}{4}).$$
The maximum height difference between any two vertices is $\sqrt{2} \lt 2$ so a monochromatic square is possible only if the height difference between some two consecutive vertices is strictly greater than $1$.
But that can't possibly happen. It obviously can't happen between $1$ and $2$, $3$ and $4$, $2$ and $4$, or $3$ and $1$. And $y_3-y_2= \cos \theta - \sin \theta = \sqrt{2} \cos (\theta + \frac{\pi}{4})$ so $0 \leq \theta \leq \frac{\pi}{2} \Rightarrow |y_3-y_2| \leq 1$, completing the proof.
Edited to respond to comments requesting additional detail.
The complete graph $K_{14}$ is much bigger than needed for this result; here are three proper subgraphs with the same property.
I. Every $2$-coloring of the complete bipartite graph $K_{3,7}$ contains a monochromatic $C_4$.
Let $K_{3,7}$ have partite sets $V_1,V_2$ with $|V_1|=3$ and $|V_2|=7$. Each vertex in $V_2$ is joined by edges of one color to two vertices in $V_1$. By the pigeonhole principle, two vertices in $V_2$ are joined by edges of the same color to the same two vertices in $V_1$.
II. Every $2$-coloring of the complete graph $K_6$ contains a monochromatic $C_4$. (This was shown in Misha Lavrov's answer; the following argument is perhaps slightly simpler.)
We may assume that there is a vertex $x$ which is incident with at most two blue edges. In fact, we may assume that $x$ is incident with exactly two blue edges; otherwise $x$ would be incident with four red edges, call them $xa_1$, $xa_2$, $xa_3$, $xa_4$, and then we could consider a vertex $y\notin\{x,a_1,a_2,a_3,a_4\}$ and proceed as in paw88789's answer. (It may be even easier to observe that the subgraph induced by $\{a_1,a_2,a_3,a_4\}$ contains either a red $P_3$ or a blue $C_4$.)
So let $x$ be incident with exactly two blue edges, $xy$ and $xz$. If one of the remaining three vertices is joined by blue to both $y$ and $z$, then we have a blue $C_4$. On the other hand, if each of those three vertices is joined by red to a vertex in $\{y,z\}$, then two of them are joined by red to the same vertex in $\{y,z\}$, and also to $x$, making a red $C_4$.
III. Every $2$-coloring of the complete bipartite graph $K_{5,5}$ contains a monochromatic $C_4$.
The proof is left as an exercise for the reader.
Best Answer
Molina, Oza, and Puttagunta, Sane bounds on Van der Waerden-type numbers, write,
The following is known: no matter how the lattice points of a coordinate plane are colored red and blue, there exists a square whose corners are the same color (a monochromatic square). In fact, using more than just two colors will still guarantee a monochromatic square (one whose vertices are the same color). So for all $c$ (the number of colors), there is a number $G(c)$ where all colorings with $c$ colors of the lattice points of a $G(c) \times G(c)$ grid will contain a monochromatic square. Unfortunately, the necessary number of points is unknown, but bounds are known. These bounds are enormous (roughly $G(c)\le2^{2^c}\times2^{2^{2^{2^{2c}}}}$).
The references they cite are
W. Gasarch, C. Kruskal, and A. Parrish. Van der Waerden’s theorem: Variants and applications. [But the Chapter that is supposed to be about "The Square Theorem" isn't there]
R. Graham, B. Rothchild, and J. Spencer. Ramsey Theory. Wiley, 1990.
B. Landmann and A. Robertson. Ramsey Theory over the integers. AMS, 2003.