Solution: Since $A$ is a subset $\mathbb{R} ^2$ (with the Euclidean topology), compactness is equivalent to closed and bounded. We will show that the set $A$ is closed and bounded in $\mathbb{R} ^2$. Note that \begin{align*} x^4 \leq x^4 + (y-1)^2 \leq 1 & \implies \vert x \vert \leq 1 \\ (y - 1)^2 \leq x^4 + (y - 1)^2 & \implies \vert y - 1 \vert \leq 1 \\ & \implies \vert y \vert \leq \vert y - 1 \vert + 1 \leq 2. \end{align*} Thus, the set $A$ is bounded.

For showing that $A$ is closed, consider the function \begin{align*} f : \mathbb{R}^2 \to \mathbb{R}, \quad (x,y) \mapsto x^4 + (y-1)^2. \end{align*} It is clear that the function $f$ is a continuous function (as it is a polynomial). Now note that \[ A = f^{-1} ((-\infty , 1]). \] Since the set $(-\infty , 1]$ is closed in $\mathbb{R} $, and $f$ is continuous, the set $A$ will be closed. Since $A$ is both closed and bounded, we conclude that $A$ is compact.