I write the Fourier transform as
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} $$
Consider, rather, the integral
$$ \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i x}-e^{-i x}}{x} e^{i k x} $$
$$ = \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} - \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} $$
Consider the following integral corresponding to the first integral:
$$\oint_C dz \: \frac{e^{i (1+k) z}}{z} $$
where $C$ is the contour defined in the illustration below:
This integral is zero because there are no poles contained within the contour. Write the integral over the various pieces of the contour:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} + \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} + \int_{-R}^{-r} dx \: \frac{e^{i (1+k) x}}{x} + \int_{r}^{R} dx \: \frac{e^{i (1+k) x}}{x} $$
Consider the first part of this integral about $C_R$, the large semicircle of radius $R$:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} = i \int_0^{\pi} d \theta e^{i (1+k) R (\cos{\theta} + i \sin{\theta})} $$
$$ = i \int_0^{\pi} d \theta e^{i (1+k) R \cos{\theta}} e^{-(1+k) R \sin{\theta}} $$
By Jordan's lemma, this integral vanishes as $R \rightarrow \infty$ when $1+k > 0$. On the other hand,
$$ \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} = i \int_{\pi}^0 d \phi \: e^{i (1+k) r e^{i \phi}} $$
This integral takes the value $-i \pi$ as $r \rightarrow 0$. We may then say that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = i \pi & 1+k > 0\\ \end{align}$$
When $1+k < 0$, Jordan's lemma does not apply, and we need to use another contour. A contour for which Jordan's lemma does apply is one flipped about the $\Re{z}=x$ axis. By using similar steps as above, it is straightforward to show that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = -i \pi & 1+k < 0\\ \end{align}$$
Using a similar analysis as above, we find that
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} = \begin{cases} -i \pi & 1-k < 0 \\ i \pi & 1-k >0 \\ \end{cases} $$
We may now say that
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} = \begin{cases} \pi & |k| < 1 \\ 0 & |k| > 1 \\ \end{cases} $$
To translate to your definition of the FT, divide the RHS by $\sqrt{2 \pi}$.
Suppose that both $f$ and $g$ are in $L^{1}$. Then $\hat{f}$ and $\hat{g}$ are bounded and, therefore, $\hat{f}g$, $f\hat{g}$ are in $L^{1}$. And,
\begin{align}
\int_{-\infty}^{\infty}\hat{f}(s)g(s)ds & =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-ist}dt g(s)ds \\
& = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\int_{-\infty}^{\infty}e^{-ist}g(s)ds \\
& = \int_{-\infty}^{\infty}f(t)\hat{g}(t)dt.
\end{align}
Suppose $f,g \in \mathcal{C}_{c}^{\infty}(\mathbb{R})$ (i.e., are compactly supported $C^{\infty}$ functions.) Then $f,g,\mathcal{F}f,\mathcal{F}g$ are in $L^{1}$, and the above gives
$$
\int_{-\infty}^{\infty}\mathcal{F}f\overline{\mathcal{F}g}ds=
\int_{-\infty}^{\infty}\mathcal{F}f\mathcal{F^{-1}}\overline{g}ds
= \int_{-\infty}^{\infty}f\mathcal{F}\mathcal{F^{-1}}\overline{g}dt
= \int_{-\infty}^{\infty}f\overline{g}dt
$$
Therefore, $\|\mathcal{F}f\|=\|f\|$. Replace $g$ by $\mathcal{F}^{-1}g$ in the above:
$$
(\mathcal{F}f,g)=(f,\mathcal{F}^{-1}g).
$$
Therefore $\mathcal{F}$, $\mathcal{F}^{-1}$ extend by continuity to isometries on $L^{2}(\mathbb{R})$, and the last equality similarly extends, which gives $\mathcal{F}^{\star}=\mathcal{F}^{-1}$ and $(\mathcal{F}^{-1})^{\star}=\mathcal{F}$. So $\mathcal{F}$ is unitary.
Best Answer
Claim: Let $T$ be an invertible linear transformation. Then $T^{-1}$ is a linear transformation
Proof: $$av+w=T^{-1}T(av+w)=T^{-1}(aT(v)+bT(w))$$ Now write $v'=T(v)$ and $w'=T(w)$. We get $$aT^{-1}(v')+bT^{-1}(w')=T^{-1}(av'+bw')$$