Take two closed disks as subsets of $\Bbb R^2$ such that they intersect at exactly one point. Let $\Bbb R^2$ have the standard euclidean topology $\mathcal J_E$ and give the above set the subspace topology.
Is this space: $a)$ path connected or $b)$ connected?
Now I would assume this is path connected, like I can surely define a path from one disk to the other through the point of intersection continuously? The disks themselves are convex so that sorts them individually. And then path connected implies connected.
Apparently I'm supposed to pay attention to the word continuous...
Lets suppose your two disks are $D_1, D_2$ and assume $x\in D_1\cap D_2$. Then let $d_1\in D_1$ and $d_2\in D_2$. We know that since each disk is connected there exists a continuous function $f:[0,1]\rightarrow D_1$ so that $f(0)=d_1$ and $f(1)=x$. And similarly there is a continuous function $g:[1,2]\rightarrow D_2$ such that $g(1)=x$ and $g(2)=d_2$. If you glue the functions $f,g$ together what can you say about the continuity of the glued function?
To show connectedness you do something similar. Suppose our space is path connected but not connected. Thus we may write $D=D_1\cup D_2$ as the union of two proper clopen subspaces, say $A_1$ and $A_2$. So in other words $D=A_1\cup A_2$ where $A_1,A_2$ are clopen in the subspace topology. Now take $a_1\in A_1$ and $a_2\in A_2$. Since our space D is path connected there is a continuous function $f:[0,1]\rightarrow D$ such that $f(0)=a_1$ and $f(1)=a_2$. But note that $f^{pre}(A_1)$ and $f^{pre}(A_2)$ are clopen. What's the contradiction?