= 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 \b[path connectedness implies connectedness].
For the specified subspace, first consider $Y=I\cup\bigcup_{n\ge1}J_n$. 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 \b[$\boxed{X\text{ is connected}}$]. This uses density, not an unproved path from the added vertical segment to the teeth.
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