Show that the dumbbell O-O (where there's no space between the "O" and "-") and the letter $\theta$ are homotopy equivalent, using the definition.
So, let $X$ be the set of points in the dumbbell, and $Y$ the set of points in $\theta$. We should give continuous maps $f:X\rightarrow Y$ and $g:Y\rightarrow X$ such that $f\circ g$ is homotopic to $\text{id}_Y$ and $g\circ f$ is homotopic to $\text{id}_X$.
I'm thinking about mapping the "-" in the dumbbell to the $-$ in $\theta$, and the two "O"s in the dumbbell to the two halves of $\theta$. But each end of the "-" in the dumbbell is connected to only one half, while each end of the $-$ in theta is connected to both halves. So I'm not sure what to do.
Further to Léo's comment, I will use an intermediate step; consider the unit disk $D^2$ with two smaller disks removed. I will show the three graphs (figure eight, dumbell, theta) will all be homotopy equivalent to this space. Writing $B(\varepsilon, x)$ for the open ball of radius $\varepsilon$ based at $x$, define $$D=D^2-B(1/4,1/2)-B(1/4,-1/2).$$
Notice that $D$ deformation retracts onto the three aforementioned graphs, $G$, if we position them appropriately (i.e. as subsets of $D$ with the points $\pm1/2$ inside the "loops" of the dumbell, figure eight, and theta). So there exists a homotopy $f_t\colon D\to D$ such that $f_0=id_D$, $f_1(D)=G$, and $f_t\restriction_G=G$ for all $t$. Let $i\colon G\to D$ be the inclusion; now, $f_1\circ i=id_G$, and $i\circ f_1^\prime =f_1$ (where $f_1^\prime$ is $f_1$ viewed as a map $f_1\colon D\to G$). Of course $id_G\simeq id_G$, and $f_1\simeq id_D$ by the homotopy $f_t$.
All three graphs are homotopy equivalent to $D$, and hence to eachother by transitivity of $\simeq$.