Let $(E, \|\cdot\|)$ be a real normed space, and let $(A_n)_{n \in \mathbb{N}}$ be a decreasing sequence of bounded closed subsets of $E$. Assume moreover that there is a real $r>0$ such that for all $n \in \mathbb{N}$, $\mathrm{diam}(A_n) > r$.
Do we have then $\mathrm{diam} \Big( \bigcap_{n \in \mathbb{N}} A_n \Big) \geq r$ ?
The answer is no.
Consider the normed space $\ell^2$ and the canonical basis $(e_n)_n$.
Define $A_n = \{e_n, e_{n+1}, \ldots\}$ for $n \in \mathbb{N}$. Clearly $(A_n)_n$ is a decreasing sequence.
We have $\|e_n\|_2 = 1, \forall n\in \mathbb{N}$ so $A_n$ are bounded. Also we have $\|e_n - e_m\|_2 = 1, \forall m \ne n$ so $A_n$ are discrete and hence closed. We see that $\operatorname{diam} A_n = 1> \frac12$ for all $n \in \mathbb{N}$.
However, $\bigcap_{n\in\mathbb{N}} A_n = \emptyset$ so $\operatorname{diam} \bigcap_{n\in\mathbb{N}} A_n = 0$ which is not $\ge \frac12$.
This construction can be done in any infinite-dimensional normed space using the Riesz lemma.
The statement is still seems false if we require $A_n$ to be convex sets: just take the closure of the convex hull $\overline{\operatorname{conv} A_n}$ of the sets defined above. The resulting sequence is still a decreasing sequence of closed bounded sets (as they are contained in the unit ball of $\ell^2$) which are now convex, and with diameter $> \frac12$ but we still have $\bigcap_{n\in\mathbb{N}} \overline{\operatorname{conv} A_n} = \emptyset$.
However, if we require the space $E$ to be finite-dimensional, then the statement is true:
For each $n \in \mathbb{N}$ since $\operatorname{diam} A_n > r$ we can choose $x_n, y_n \in A_n$ such that $d(x_n, y_n) > r$.
$A_1$ is a closed and bounded set and hence compact (by finite-dimensionality). Hence the sequences $(x_n)_n$ and $(y_n)_n$ in $A_1$ have limit points in $A_1$, i.e. there exist subsequences $(x_{p(n)})_n$ and $(y_{p(n)})_n$ such that $x_{p(n)} \to x$ and $y_{p(n)} \to y$ with $x,y \in A_1$.
Let $m \in \mathbb{N}$ and notice that $\left(x_{p(n)}\right)_{n=m}^\infty$ and $\left(y_{p(n)}\right)_{n=m}^\infty$ are sequences in $A_m$ which clearly converge to $x$ and $y$ respectively. Since $A_m$ is closed we have $x,y \in A_m$. Therefore, $x,y \in \bigcap_{n\in\mathbb{N}} A_n$ and by the continuity of the metric
$$\underbrace{d\left(x_{p(n)}, y_{p(n)}\right)}_{> r} \xrightarrow{n\to\infty} d(x,y)$$
We conclude $d(x,y) \ge r$ so $\operatorname{diam} \bigcap_{n\in\mathbb{N}} A_n \ge r$.