Curious to see whether my proof below is acceptable or not. Any feedback will be well received.
Many thanks.
Every graph G contains a path with $\delta(G)$ edges.
$\mathbf{Proof.}$ Let $G$ be a graph with $\delta(G)=d$, and vertex set $\{v_{1},v_{2},v_{3},...,v_{n}\}$. Now consider the neighbourhood of a vertex $v_{i}$ with degree $d$. Each distinct vertex incident to $v_{i}$ will have a degree greater than or equal to $d$. This implies that graph $G$ has at least $d+1$ vertices, one with degree $d$ and the remaining vertices with degree at least $d$. $\delta(G)$ forces graph $G$ to be connected. Because there are at least $d+1$ vertices, there will exist a path with $d+1-1=d$ edges.