# How to show $\text{diam}\left(\bigcup\limits_{i=1}^{\infty}M_i\right)\leq \sum \limits_{i=1}^{\infty}\text{diam}(M_i)$

Let $$(X,d)$$ be a metric space. If $$M_i\subset X$$ with $$M_i \cap M_{i+1}\neq \emptyset$$ for $$i\in \mathbb{N}$$, then:$$\text{diam}\left(\bigcup\limits_{i=1}^{\infty}M_i\right)\leq \sum \limits_{i=1}^{\infty}\text{diam}(M_i)$$ I feel like this proof needs the triangle inequality, but I can't get ahead of this problem - any suggestions on how to prove it?

Take $$x \in M_i, y \in M_j$$ with $$i \le j$$. Let $$x_i=x$$ and $$x_j = y$$
Choose $$x_k \in M_k \cap M_{k+1}$$ for $$k=i,...,j-1$$.
Then $$d(x,y) = d(x_i,x_j) \le \sum_{k=i}^{j-1} d(x_k,x_{k+1}) \le \sum_{k=i}^{j-1} \operatorname{diam} M_k \le \sum_{k=1}^{\infty} \operatorname{diam} M_k$$.
Hence $$\operatorname{diam} \cup_k M_k \le \sum_k \operatorname{diam} M_k$$.