Question about open set and a curve homopotic to a point If image of a simple loop is contained in an open connected set $D\subseteq R^2$ such that this loop is homotopic to a point in $D$, will interior of curve be contained in $D$.
What if we take closed contour instead of simple loop?
Definition: Simple loop is homeomorphic to $S^1$.
A similar question: Inside of boundary of an open region
 A: Yes to the first question. Let $\gamma : S^1 \to \mathbb{R}^2$ be the curve. Suppose that a point of the interior of the curve, say $x$, is not in $D$. Then $a = \gamma_*[S^1]$, the pushforward of the fundamental class of $S^1$, defines a generator of the homology group $H_1(\mathbb{R}^2 \setminus \{x\}) = \mathbb{Z}$ – see the proof of the Jordan–Brouwer theorem. However, the fact that $\gamma$ is nullhomotopic in $D$ shows that $b = \gamma_*[S^1] = 0$ in $H_1(D)$. (I'm abusing notation a bit by writing the same $\gamma$ but with different codomains, either $D$ or $\mathbb{R}^2 \setminus \{x\}$).
But we have an inclusion $i : D \to \mathbb{R}^2 \setminus \{x\}$ (recall that $x \not\in D$), and functoriality of homology then tells us that $b \in H_1(D)$ is sent to $a \in H_1(\mathbb{R}^2 \setminus \{x\})$, i.e. $i_* b = a$. But we know that $b = 0$ while $a \neq 0$. This is a contradiction.
A: You may know this, but here is how we talk about homotopy formally:
Two maps $f,g: X \to Y$ are homotopic if there is a map continuous $H: X \times [0,1] \to Y$ such that $f(x) = H(x,0)$ and $g(x) = H(x,1)$ for all $x \in X$. You can think of the $[0,1]$ here as being time, so that the first map is turning into the second map.
In your case $f: S^1 \to D$ and $g: S^1 \to D$ where $g$ is constant (say $g(x) = p$ for all $x \in S^1$. Here, $p$ is that point). The statment is then that there is a continuous $H: S^1 \times [0,1] \to D$ such that $f(x) = H(x,0)$ and $g(x) = H(x,1) = p$ for all $x \in S^1$. 
So here is an idea for a simpler proof: Suppose it does not contain all of the "interior" points. Consider one such point $x$. Let $T$ be the set of all $t$ such that image of $X$ at $t$, $H(X,t)$ contains the $x$ as an "interior" point. Define $T_0$ to be the set of $t \in T$ such that for all $u < t$, $u \in T$. That is, $T_0 = \{t \in T: (\forall u < t)(u \in T)\}$. We know that this set is nonempty since $0 \in T$. Also, note that $1 \not \in T$. Then consider $\sup(T_0)$ and get a contradiction that $H(X,\sup(T_0))$ cannot have the point either inside or outside of its "interior" (i.e. $\sup(T_0) \not \in T$  and $\sup(T_0) \in T$).
Since $\{x\}$ and $H(X,\sup(T_0))$ are compact, there exists and $\eta>0$
such that $D(x,\eta) \cap H(X,\sup(T_0)) = \emptyset$. So $D(x,\frac{\eta}{2}) \cap \displaystyle\bigcup_{y \in H(X,\sup(T_0))}D(y,\frac{\eta}{2}) = \emptyset$. Using the uniform continuity of $H$ (since $S^1 \times [0,1]$ is compact), we get a $\delta > 0$ such that $(\forall y,z \in S^1 \times [0,1]:d(y,z)<\delta)(d(H(y),H(z))<\eta)$. Using this, you can show that for $t \in (\sup(T_0) - \delta,\sup(T_0) + \delta)$, $x$ is in the "interior" of $H(X,t)$ iff $x$ is in the "interior" of $H(X,\sup(T_0))$.
(Here, we think of the "interior" of a curve as the union of all of the bounded connected components of $\mathbb{R}^2$\curve)
