For some details missing from the derivation below, see my answer to this question.
First, we write the LHS in terms of its components, using the Kronecker delta and Levi-Civita symbols. (Note: I'll drop the vector arrow on $\vec{A}$ but it's a vector)
$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\partial_{j}\,(\nabla \times A)_{k} = \epsilon_{ijk}\,\partial_{j}\,(\epsilon_{krs}\,\partial_{r}\,A_{s})
$$
The $\epsilon_{krs}$ are constants, so
$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\epsilon_{krs}\,\partial_{j}\,\partial_{r}\,A_{s}
$$
But
$$\epsilon_{ijk}\,\epsilon_{krs} =
\delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr}
$$
so
$$[\,\nabla \times (\nabla \times A)\,]_{i} =
(\delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr})\,
\partial_{j}\,\partial_{r}\,A_{s} =
\delta_{ir}\,\delta_{js}\,\partial_{j}\,\partial_{r}\,A_{s} -
\delta_{is}\,\delta_{jr}\,\partial_{j}\,\partial_{r}\,A_{s}
$$
Simplifying,
$$[\,\nabla \times (\nabla \times A)\,]_{i} =
\partial_{j}\,\partial_{i}\,A_{j} -
\partial_{j}\,\partial_{j}\,A_{i} =
\partial_{i}\,(\partial_{j}\,A_{j}) -
(\partial_{j}\,\partial_{j})\,A_{i} =
\partial_{i}\,(\nabla \cdot A) - \nabla^2A_i
$$
so
$$\nabla \times (\nabla \times A) =
\nabla(\nabla \cdot A) - \nabla^2A
$$
Lucky you, the appearance of a totally wrong answer persuaded me to post a complete solution.
Fundamental Theorem of Calculus for line integrals: Suppose $\Omega\subset\Bbb R^2$ is open and $f\in C^1(\Omega)$. If $\gamma:[0,1]\to\Omega$ is a $C^1$ curve then $\int_\gamma\nabla f\cdot dr=f(\gamma(1))-f(\gamma(0))$.
(Hence if $\gamma$ is a closed curve (meaning $\gamma(1)=\gamma(0)$) the integral is $0$. With $\Omega=\Bbb R^2\setminus\{(0,0)\}$ and $\gamma(t)=(\cos(2\pi t),\sin(2\pi t))$ this proves the result you ask about; in particular what happens at the origin doesn't matter.)
Proof, even though it must be in the book: By definition, if $F$ is a vector field then $$\int_\gamma F\cdot dr=\int_0^1F(\gamma(t))\cdot\gamma'(t)\,dt.$$Now suppose that $F=\nabla f$, and set $$g(t)=f(\gamma(t)).$$The chain rule shows that $$F(\gamma(t))\cdot \gamma'(t)=g'(t),$$so $$\int_\gamma F\cdot dr=\int_0^1g'(t)\,dt=g(1)-g(0).$$
Corollary: If $\Omega=\Bbb R^2\setminus\{(0,0)\}$ and $F$ is the infamous vector field mentioned in the question there does not exist $f\in C^1(\Omega)$ with $F=\nabla f$.
(Indeed, in the language used in calculus books, $F$ is the standard example of a vector field which is closed but not exact.)
Best Answer
The statement is true when $\gamma$ is a positively oriented simple closed curve bounding some Jordan domain $\Omega$.
Let $\mathcal{I}$ be the integral at hand. Identify Euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$ and introduce complex coordinates $x,y$:
$$ \begin{cases} \vec{x} = (x_1,x_2) & \leftrightarrow & x = x_1 + x_2 i\\ \vec{y} = (y_1,y_2) &\leftrightarrow & y = y_1 + y_2 i \end{cases}$$ Let $z = y - x$. For any $\vec{\rho} = (\rho_1,\rho_2) \leftrightarrow \rho = \rho_1 + i\rho_2 \in \mathbb{C}$, we will use the notation $\Omega + \rho$ and $\gamma + \rho$ to denote the image of $\Omega$ and $\gamma$ under translation $\vec{\rho}$.
In terms of complex coordinates, we have $d\vec{x} \cdot d\vec{y} = \frac12 \left(dx d\bar{y} + dy d\bar{x}\right)$. This leads to
$$\begin{align} \mathcal{I} &= -\frac{1}{8\pi} \int_{\gamma \times \gamma} \log(z\bar{z}) (dx d\bar{y} + dyd\bar{x}) = -\frac{1}{4\pi}\Re \left[\int_{\gamma\times\gamma} \log(z\bar{z}) dx d\bar{y}\right]\\ &= -\frac{1}{4\pi}\Re\left[\int_{x \in \gamma} \left(\int_{z \in \gamma - x}\log(z\bar{z}) d\bar{z}\right) dx\right] \end{align}\tag{*1} $$ Apply Stoke's theorem (the version for complex coordinates) to the inner integral, we obtain $$\begin{align} \mathcal{I} &= -\frac{1}{4\pi}\Re\left[\int_{x\in\gamma} \left(\int_{z \in \Omega - x} \left(\frac{dz}{z} + \frac{d\bar{z}}{\bar{z}}\right)\wedge d\bar{z}\right) dx\right]\\ &= \frac{1}{4\pi}\Re\left[\int_{x\in\gamma} \left( \int_{z\in \Omega - x}\frac{d\bar{z} \wedge dz}{z} \right) dx\right]\\ &= \frac{1}{4\pi} \Re\left[\int_{x\in\gamma} \left( \int_{y\in \Omega}\frac{d\bar{y} \wedge dy}{y-x} \right) dx\right]\\ &= \frac{1}{4\pi} \Re\left[ \int_{y\in\Omega}\left(\int_{x\in\gamma}\frac{dx}{y-x}\right) d\bar{y} \wedge dy \right] \end{align} $$ By Cauchy's integral formula, we have $$\int_{x\in\gamma}\frac{dx}{y-x} = -2\pi i\quad\text{ for } y \in \Omega$$ As a result, $$\begin{align}\mathcal{I} &= \frac{1}{4\pi} \Re\left[ \int_{y \in \Omega} (-2\pi i)(2i dy_1 \wedge dy_2 )\right] = \Re\left[ \int_{y \in \Omega} dy_1 \wedge dy_2 \right]\\ &= \Re\left[\verb/Area/(\Omega)\right] = \verb/Area/(\Omega) \end{align} $$
Update - Generalization to non-simple closed curves.
For non-simple closed curve $\gamma$ and $y \not\in \gamma$, let $W(\gamma;y)$ be the winding number of $\gamma$ around $y$. We have following generalization of Cauchy integral formula:
$$\int_\gamma \frac{dx}{y-x} = -2\pi i W(\gamma;y)\tag{*2}$$
When $\gamma$ satisfies following conditions:
We can break any contour integral over $\gamma$ to integral combinations of contour integrals over boundaries of connected components in step $2$. We can apply Stoke's theorem to the individual components and recombine the results.
Apply this procedure to inner contour integral in $(*1)$ and with help of $(*2)$, we obtain:
$$ \mathcal{I} = \frac{1}{4\pi} \Re\left[\int_{x\in\gamma} \left( \int_{y\in \mathbb{C}\setminus\gamma}\frac{W(\gamma;y)}{y-x} d\bar{y} \wedge dy \right) dx\right] = \int_{y\in \mathbb{C}\setminus\gamma} W(\gamma;y)^2 dy_1 \wedge dy_2$$
Recall winding number is constant over each connected component. If $\Omega_1, \ldots, \Omega_m$ are the connected components of $\mathbb{C}\setminus\gamma$ with non-zero winding numbers and $W_i$ is the winding number for $\Omega_i$, we can rewrite last expression as
$$\mathcal{I} = \sum_{i=1}^m W_i^2 \verb/Area/(\Omega_i)$$
The integral $\mathcal{I}$ is simply a weighted sum of the areas of connected components and the weight equals to the square of corresponding winding number.
In the special case where all $|W_i| \le 1$, above formula simplifies to
$$\mathcal{I} = \sum_{i=1}^m \verb/Area/(\Omega_i)$$
$\mathcal{I}$ becomes the unsigned area of those connected components whose winding number is non-zero.