Suppose a simply connected domain D consists of two disjoint domains $D_1$ and $D_2$ and a curve r between them ($D_1$ and $D_2$ share a common boundary piece r); the function $f(z)$ is analytic in $D_1$ and $D_2$ and continuous in D. Show that f(z) is analytic in D.
Wouldn't a counterexample be the function $e^{x−\frac{1}{x}}$ with D being the disk of radius 1 about 1? Then at the point (0,0) the function wouldn't be defined... But I wonder would that break the continuity condition? I am confused about how to solve this if it ins't true that if $f(z)$ is analytic in a simply connected domain $D$ and continuous in $\bar{D}$, then $\oint_{\delta D}f(z)dz=0$.
Morera's theorem is the standard tool to prove this.
In your attempted counterexample, the function is indeed not continuous at $0$, and can't be extended to be continuous there, because it is unbounded as $z \to 0$. In fact, if an analytic function is bounded in a neighbourhood of an isolated singularity, that singularity is removable.