Let $X$ be the letter

$$\ \ \ \ \ \mathsf{X}\ \ \ \ \ $$
and $Y$ be the letter

$$\ \ \ \ \ \mathsf{Y}\ \ \ \ \ $$

Then $X$ and $Y$ are homotopy-equivalent, but they are not homeomorphic.

**Sketch proof:** let $f:X\to Y$ map three of the prongs of the $\mathsf{X}$ on to the $\mathsf{Y}$ in the obvious way, and let it map the fourth prong to the point at the centre. Let $g:Y\to X$ map the $\mathsf{Y}$ into those three prongs of the $\mathsf{X}$. Then $f$ and $g$ are both continuous, and $f$ is a surjection but is not injective, while $g$ is an injection but is not surjective. Now the compositions $f\circ g$ and $g\circ f$ are both easily seen to be homotopic to the identities on $X$ and $Y$, so $X$ and $Y$ are homotopy-equivalent.

In other words, observe that $\mathsf Y$ is a *deformation retract* of $\mathsf X$. Alternatively, observe that $\mathsf X$ and $\mathsf Y$ both retract on to the point at the centre.

On the other hand, $X$ and $Y$ are not homeomorphic. For example, removing the point at the centre of the $\mathsf{X}$ yields a space with four connected components, while removing any point from the $\mathsf{Y}$ yields at most three connected components.