Topologist's sine curve

Counterexample

Let $$T = \left\{ \left( x, \sin \tfrac{1}{x} \right) : 0 < x \le 1 \right\} \cup \{ (0, 0) \} \subseteq \mathbb{R}^2,$$ with the subspace topology of the plane. Then $T$ is connected but not path-connected. Thus connected spaces need not be path-connected.

Remarks

The graph part $G$ is connected, as the continuous image of $(0, 1]$. The origin is in the closure of $G$, and a set between a connected set and its closure is connected. Thus $T$ is connected. No path joins the origin to a point of $G$. Near $x = 0$, the curve goes up and down between $-1$ and $1$ infinitely often, and a path cannot follow it continuously. Some texts use the closed version, which adds the whole segment $\{0\} \times [-1, 1]$. It is compact, connected and not path-connected. The space $T$ is also not locally connected.

Explanations

L1 human

The curve $\sin(1/x)$ oscillates faster and faster as $x$ goes to $0$. The origin is a limit of the curve, so the space is in one piece. But you cannot walk along the curve to the origin, because the curve never settles down.