Use spherical coordinates $\phi$, $\theta$, where $\theta=0$ on the equator, and produce corresponding points on the sphere $S^2$ and the cylinder $C$ as follows:
$$\eqalign{(\phi,\theta)\mapsto{\bf r}(\phi,\theta)&:=(\cos\phi\cos\theta,\sin\phi\cos\theta, \sin\theta)\ ,\cr
(\phi,\theta)\mapsto{\bf f}(\phi,\theta)&:=(\cos\phi,\sin\phi, \sin\theta)\ .\cr}$$
In this way Lambert's projection $L$ appears as
$$L:\ S^2\to C\ ,\qquad {\bf r}(\phi,\theta)\to{\bf f}(\phi,\theta)\ .$$
Then compute
$$\eqalign{{\bf r}_\phi&=(-\sin\phi\cos\theta,\cos\phi\cos\theta,0),\quad {\bf r}_\theta=(-\cos\phi\sin\theta,-\sin\phi\sin\theta,\cos\theta),\cr
{\bf f}_\phi&=(-\sin\phi,\cos\phi,0),\quad {\bf f}_\theta=(0,0,\cos\theta)\ .\cr}$$
The fact that $$|{\bf r}_\phi|=\cos\theta\ne 1=|{\bf f}_\phi|\qquad(\theta\ne0)\tag{1}$$
already proves that $L$ is not an isometry: Equation (1) says that a piece of parallel at latitude $\theta$ on the sphere has length $$\int_{\phi_0}^{\phi_1}|{\bf r}_\phi|\ d\phi=(\phi_1-\phi_0)\cos\theta\ ,$$ while the projected arc has length $(\phi_1-\phi_0)$, given that $\phi_1>\phi_0$. Second, from
$$ {|{\bf r}_\phi|/|{\bf f}_\phi|\over |{\bf r}_\theta|/|{\bf f}_\theta|}=\cos^2\theta\ne1\qquad(\theta\ne0)\tag{2}$$
one then deduces that the projection cannot be conformal, even though the right angle between ${\bf r}_\phi$ and ${\bf r}_\theta$ is preserved. Equation $(2)$ says that at any point ${\bf p}:={\bf r}(\phi,\theta)\in S^2$ where $\theta\ne0$ the local dilatations of the parallel and the meridian through ${\bf p}$ under $L$ are different. For a conformal map the dilatations in all directions emanating from ${\bf p}$ would have to be equal.
Finally one computes
$$|{\bf r}_\phi\times{\bf r}_\theta|=\cos\theta=|{\bf f}_\phi\times{\bf f}_\theta|\ .$$
This shows that to an infinitesimal rectangle in the parameter plane correspond isoareal infinitesimal rectangles on $S^2$, resp. $C$. Therefore the projection is area-preserving.
