 Solution

A connected space admits no separation into two disjoint nonempty open subsets whose union is the space. A path-connected space has, for every pair of points, a continuous path from $[0,1]$ joining them. If a path-connected space had a separation $U,V$, choose its endpoints in different pieces. Their inverse images under the path would separate the connected interval $[0,1]$, a contradiction. Thus path connectedness implies connectedness.

For the specified subspace, first consider $Y=I\cup {\bigcup }_{n\ge 1}{J}_n$. Every point of a tooth $J_n$ can move vertically to its base in $I$, then horizontally to the origin; points already on $I$ use only the horizontal segment. Concatenating these paths shows that $Y$ is path-connected and hence connected.

Every point $(0,t)$ of $A$ is a limit of $(1/n,t)\in Y$. Consequently $Y$ is dense in $X=Y\cup A$ in the subspace topology. To prove $X$ connected, suppose $X=U\cup V$ were a separation. Restricting to the connected $Y$ puts all of $Y$ in one piece, say $U$. But the other nonempty relatively open piece $V$ must meet dense $Y$, a contradiction. Therefore $\boxed{X\text{ is connected}}$. This uses density, not an unproved path from the added vertical segment to the teeth.