Area of the part of a sphere Let $x \in \mathbb{R^3}$ and $t>0$. Also let $R>0$. Why is the area of the part of the sphere $\partial B(x,t)$ inside $B(0,R)$ smaller than the area of $\partial B(0,R)$:
$$|\partial B(x,t) \cap B(0,R)| \leq |\partial B(0,R)|$$
How can this be shown rigorously?   
 A: Let $C = B(0, R)$ and $X = B(x, t)$ where we rotated the scene to have 
$x = (d,0,0)$.

The very easy case is $C \cap X = \emptyset$ (empty intersection). This happens if $d-t > R$, because the left-most point of $X$, which is $(d-t,0,0)$ is separated from the right-most point of $C$, which is $(R,0,0)$.

The easy case is $X \subseteq C$ ($X$ fits in $C$). To fit the left- and right-most points of $X$ within $C$ we need $-R \le d-t$ and $d+t\le R$.
This means $\pm d \le R-t$, so $0 \le R-t$ and $t \le R$, and 
$$
A_X = 4\pi t^2 \le 4\pi R^2 = A_C
$$
For reversed roles $C \subseteq X$, then $C \cap X = C$ and the inequality holds equal.

Now the general case: We attempt to calculate the area within the intersection. For the area calculation we will use spherical coordinates,
for this we rotate and translate the scene such that $X$
centers around the origin and $C$ around $(0,0,d)$.

We need the upper bound for the polar angle $\theta_i$.
For points in $I = \partial C\cap \partial X$ we have
$$
x^2 + y^2 + z^2 = t^2 \\
x^2 + y^2 + (z - d)^2 = R^2
$$
The difference equation is
$$
-2zd + d^2 = R^2 - t^2 \Rightarrow \\
z_i = \frac{d^2 - R^2 + t^2}{2d}
$$
where $z_i$ is the $z$-coordinate of the intersection $I$, which shows up as red line or red plane $z=z_i$ in the images above.
Then $I$ is given by
$$
x^2 + y^2 + z_i^2 = t^2
$$
which is a circle with radius $r = \sqrt{t^2-z_i^2}$.
This gives
$$
\cos\theta_i = z_i/t
$$
For the surface of $X$ within $C$ we get
$$
A = 
\int\limits_0^{2\pi}\!\!d\varphi
\int\limits_0^{\theta_i}\!\! d\theta \, t^2 \sin\theta
= 2\pi t^2 [-\cos\theta]_0^{\theta_i}
= 2\pi t^2 (1-\cos\theta_i)
= 2\pi t^2(1-z_i/t)
$$
From the sphere equation for $X$ we have $\lvert z\rvert \le t \iff z\in [-t,t]$, which limits $z_i$ as well and this means 
$$
0 \le A \le 4\pi t^2 = A_X
$$
as expected. Now we need to compare $t$ and $R$. 
If $t\le R$ we have $A \le A_X \le 4\pi R^2 = A_C$ and we are done.
A: If $t<R$ the result is obvious, let's then suppose $t>R$ and let the two spherical surfaces intersect along a circle of radius $r$ (if they don't intersect the result is once again trivial): the part of the larger sphere contained inside the smaller sphere is a spherical cap, the smaller of the two spherical caps on the larger sphere having the said circle as base.
The area of the smaller spherical cap of base radius $r$ is given by 
$S(\rho)=2\pi\rho(\rho-\sqrt{\rho^2-r^2})$, where $\rho$ is the radius of the sphere. This is a decreasing function of $\rho$ (when $r$ is fixed), so we have $S(t)<S(R)<4\pi R^2$. 
In other words: the area of the spherical cap of the larger sphere inside the smaller sphere is less than the spherical cap of the smaller sphere, which in turn is less than the area of the smaller sphere.
A: About the question "How can this be shown rigorously?".
It has been answered by mvw and Artetino's excellent answers. 
I will like to point out something interesting about "Why is this true for spheres?". The answer is sphere is a convex body! If one replace $B(0,R)$ and $B(x,t)$ by any two convex bodies $M, N \subset \mathbb{R}^3$, one still have
$$\verb/Area/(\partial N \cap M) \le \verb/Area/(\partial M)$$
To prove this, we need a result from Cauchy.

Cauchy surface area formula
Given any convex body $K \subset \mathbb{R}^3$ and unit vector $\hat{n} \in S^2$. Let $A_K(\hat{n})$ be the area of the image of $K$ under orthogonal projection onto any plane having $\hat{n}$ as a normal vector. If one average $A_K(\hat{n})$ over all possible directions of $\hat{n}$, one get one quarter the surface area of $K$.
  $$\frac{1}{4\pi}\int_{S^2} A_K(\hat{n}) d\mu(\hat{n}) = \frac14\verb/Area/(K)$$

As a corollary of this, if $K \subset L$ are two convex bodies, then
$A_K(\hat{n}) \le A_L(\hat{n})$ for all $\hat{n} \in S^2$ and hence $\verb/Area/(K) \le \verb/Area/(L)$. Set $K = M\cap N$ and $L = M$, we immediately
obtain
$$\verb/Area/(\partial N \cap M) \le \verb/Area/(\partial(N \cap M)) \le \verb/Area/(M)$$
Apply this to our case where $M = B(0,R)$ and $N = B(x,t)$, we see the area of
$\partial B(x,t) \cap B(0,R)$ is not only smaller than that of $\partial B(0,R)$, it is smaller than that part of $\partial B(0,R)$ outside $B(x,t)$.
