Prove $B(P,r)\subseteq B(P,r')\implies r\leq r'$ 
Prove $B(P,r)\subseteq B(P,r')\implies r\leq r'$

where $B(P,r)$ is the set of points inside a circle with radius $r$ centered at $P$ and $d(P,Q)$ is the distance between $P$ and $Q$. So
$$B(P,r)=\{x\mid d(P,x)\leq r\}$$

Proof: Assume $B(P,r)=B\subseteq B'=B(P,r')$. Then the area of $B$ is less or equal than the area of $B'$, so
\begin{align}
\pi r^2&\leq \pi r'^2\\
r^2&\leq r'^2\\
r&\leq r'\qquad\blacksquare
\end{align}

But I feel like I'm cheating. Is this a valid way or is there another better way to show this?
 A: In certain cases this assertion is true. For example, suppose the points lie in $\mathbb{R}^n$ and $d$ is the usual Euclidean metric. In this case the truth of the claim is easy to prove. For example, suppose on the contrary that $r > r'$ and let $x\in B(P,r)$ such that $d(P,x) = r$ (clearly such a point exists). Then $d(P,x) = r > r'$, which implies that $x\not\in B(P,r')$, a contradiction.
However, I do not believe this claim is true in general. Consider the following counterexample. Let $X$ be a set and consider the metric space $(X,d)$, where
$$
d(x,y) = \left\{
\begin{array}{ll}
1, & \text{if}~x \neq y\\
0, & \text{if}~x = y
\end{array}\right.
$$
Let $P\in X$ and define the closed ball $B(P,r) = \{x\in X\,|\, d(P,x) \leq r\}$. It is clear that if $r \geq 1$, then $B(P,r) = X$ and if $0 < r < 1$, then $B(P,r) = \{P\}$. Thus it is possible that $B(P,r) = B(P,r') \implies B(P,r) \subset B(P,r')$ with $r > r'$. The only possible issue with this counterexample is the definition of the symbol $\subseteq$. If its intended purpose is to indicate that $B(P,r)$ is a proper subset of $B(P,r')$, then this couterexample fails.
If $\subseteq$ implies proper subset, then the claim is true. Suppose that $B(P,r)$ is a proper subset of $B(P,r')$. Then there exists an $x\in B(P,r')$ that does not belong to $B(P,r)$. Therefore, $r < d(P,x) \leq r'$.
