As soon as dfs is called from a source vertex, the source gets visited in the first step. In standard dfs algorithm, we iterate through each node and check if its visited or not. If visited we call dfs function and this ensures that the node is visited after it. If it's already visited then you move ahead without doing anything. This ensures that each node is visited for sure.

You visit every index only once in the visited array.

it doesn't, it traverses every reachable node from the source node

for example:

if we start dfs from 1 it will not visit 3

Suppose that it is not the case that all vertices reachable from the source of the dfs are visited by the dfs. Then there is at least one vertex such that it is reachable from the source $$$s$$$ of the dfs, but it is not visited by the DFS. Let $$$v$$$ be such a vertex with the minimum distance to the source.

Clearly, $$$v$$$ is not $$$s$$$, since when you call dfs on $$$s$$$, the first thing you do is to mark it as visited. So the distance from $$$s$$$ to $$$v$$$ is $$$\ge 1$$$.

Since $$$v$$$ is reachable from $$$s$$$ and the distance from $$$s$$$ to $$$v$$$ must be $$$\ge 1$$$, there must be a vertex before $$$v$$$ in the path from $$$s$$$ to $$$v$$$, say $$$u$$$. By definition, $$$u$$$ is reachable from $$$s$$$, and since $$$\text{dist}(s, u) = \text{dist}(s, v) - 1 < \text{dist}(s, v)$$$, $$$u$$$ must be visited by the dfs at some point (otherwise it would contradict the choice of $$$v$$$).

Since $$$u$$$ is visited, there must have been a call to dfs with argument as $$$u$$$. There are two cases when $$$v$$$ is encountered in the adjacency list of $$$u$$$ in such a call:

• $$$v$$$ is marked as visited: this implies that $$$v$$$ has been visited by the dfs at this point, which is a contradiction to the choice of $$$v$$$.
• $$$v$$$ is not marked as visited: this implies that dfs will be called with argument $$$v$$$, so $$$v$$$ will be marked as visited immediately, which is a contradiction to the choice of $$$v$$$.

Hence the existence of such a $$$v$$$ leads to a contradiction, so dfs traverses all vertices that are reachable from the source.