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Well, the exercise it's as the title says. I know that if $G$ is connected then for every pair of vertex $u,v$ in $G$ there's a walk between them. So when the Line Graph $L(G)$ is constructed those edges in $G$ where the walk between $u$ and $v$ is formed will be adjacent vertex in the Line Graph $L(G)$. Since $G$ is connected therefore $L(G)$ must be connected.

I'm not sure about if the argument is valid, also I'm looking for a more formal/detailed proof, because I think there's a lot of holes in this argument.

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The argument given is correct, though a bit informal. Here's an elaboration:

To be more precise, suppose $G$ is connected. Then for any two vertices $u,v\in G$, there is a path $P=u\to x_1\to x_2\to\cdots\to x_n\to v$ where the $x_i$'s are some other vertices of $G$.

By the existence of such a path $P$, we can say that $u$ is adjacent to $x_1$, $x_i$ is adjacent to $x_{i+1}$ (for $1\leq i\lt n$) and $x_n$ is adjacent to $v$

By the definition of the line graph $L(G)$, this means $(u,x_1),(x_i,x_{i+1})$ and $(x_n,v)$ are vertices of $L(G)$ [for $1\leq i\lt n$]

Now, since $(u,x_1)$ and $(x_1,x_2)$ share a common endpoint $x_1$, they must be adjacent in $L(G)$. Similarly, $(x_i,x_{i+1})$ is adjacent to $(x_{i+1},x_{i+2})$ for $1\leq i\lt n-1$ and $(x_{n-1},x_n)$ is adjacent to $L(G)$

So, we have a path $L_P$ in $L(G)$ as $L_p=(u,x_1)\to (x_1,x_2)\to\cdots\to(x_{n-1},x_n)\to(x_n,v)$

Hence, for every path $P$ in $G$, there is a path $L_P$ in $L(G)$.

Suppose $(a,b)$ and $(A,B)$ are two vertices of $L(G)$. We show that there is a path between them: By definition of $L(G)$, all four of $a,b,A,B$ are vertices of $G$ with $a$ adjacent to $b$, ie, $a\to b$ and $A$ adjacent to $B$, ie, $A\to B$; since $G$ is connected, there must be a path $P_{bA}$ from $b$ to $A$ which translates to a path $L_{\large P_{bA}}$ in $L(G)$ which gives us a path $a\to b\to P_{bA}\to A\to B$ in $G$ and a path $(a,b)\to L_{\large P_{bA}}\to (A,B)$ in $L(G)$

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