We'll prove that at least one of the following colorings is valid: All cells are white. All cells are black. White-black alternating rows. White-black alternating columns. A cell $(x,y)$ is white if and only if $2\mid x$ and $2\mid y$. A cell $(x,y)$ is black if and only if $2\mid x$ and $2\mid y$. Chessboard coloring. It's easy to see that each type of $2\times 2$ tiles is contained in exactly one of the listed colorings. Hence, by pigeonhole principle, at least one of the listed colorings doesn't contain a forbidden tile.

Choose the following $2\times 2$ tiles: 1-4. All $2\times 2$ tiles in which the upper-left cell is black and the lower-left cell is white. 5. All-black tile. 6. All-white tile. 7. $\begin{bmatrix} \text{black} & \text{white} \\ \text{black} & \text{white} \\ \end{bmatrix}$ Assume, for contradiction, that there exists an infinite table coloring which avoids forbidden tiles. Since all-white coloring is invalid, such a coloring must contain a black cell. Look at a $2\times 3$ sub-table: $\begin{bmatrix}A&B\\C&D\\E&F\\\end{bmatrix}$ and assume WLOG that $A$ is black. Hence $C$,$E$ are black as well. If $D$ is black, then $F$ is black and $C,D,E,F$ is an all-black tile, contradiction. Hence $D$ is white. Since tile (7) is forbidden, $B$ must be black. Hence $D$ is black, contradiction.