So to show this is a 2 manifold with boundary you have to show that around each point there is a neighborhood that is either homeomorphic to $D^2$ or $D^2_+= \{(x,y)\in \mathbb{R} | \,\,\,\, y\geq 0, \,\,\,\, |(x,y)|<1 \}$.
Let $X$ be the described set $X / \sim$ the quotient and $\pi$ the quotient homomorphisim.
For $x \in \pi( \text{int} \, ( X )) = \text{int} \, (X)$ we are done, this set is homeomorphic to the disk. On $\text{int} \,(X)$, $\pi$ is a homeomorphism.
For $x \in \pi( (-10, 10) \times \{1\})$ consider $\pi((-10, 10) \times [1,-1))$. Similarly for the other side.
For $x \in \pi( \{10\} \times (-1,1) )$ it is more difficult. Here we have to somehow work with the twist. Let $f: [-10,-9) \cup (9,10] \times (-1,1) / \sim \,\, \to (-1,1)^2 $ be:
$$
f(x,y) = \left\{
\begin{array}{lr}
(x-10,y) & : x \in (9,10] \\
(x+10,-y) & : x \in [-10,-9)
\end{array}
\right.
$$
I claim that this is continuous and bijective. Pulling $f$ back to $X$, i.e. considering $f \circ \pi : X \to (-1,1)^2$, it is continuous (this is the universal property of quotients). And it is bijective as $f \circ \pi$ is 1 to 1 except for the points that are identified where is is 2 to 1. But those points are identified so $f$ is 1 to 1 and onto. $f$ is also an open map, any open set in $X / \sim$ is the union of the images of an open sets from $X$ and $f \circ \pi$ is clearly an open map.
What you are trying to prove is false. Let $C_1, C_2$ be two nested round circles in the plane $E^2$ which are tangent to each other at a point $p$. Take $C=C_1\cup C_2$. The complement, $E^2 \setminus C$, consists of three components, one of them is the "exterior" of the outer circle $C_1$, the second is the "interior" of the inner circle $C_2$. Let $U$ be the third component. It is bounded, regular and simply connected, $\partial U=C$. Take any open arc $a\subset C$ containing $p$. Then $C\setminus a$ is disconnected.
Best Answer
This can all be done without recourse to the classification of surfaces or the Seifert-van Kampen theorem. Instead, I would say that a good understanding of this problem is an important step in (some treatments of) the proof of the classification of surfaces.
For (a), let's use $p : A - D \to F$ to be the quotient map. Also, let $D'$ be a larger open disc whose interior contains $D$ concentrically. Consider the boundary circles $C' = \partial D'$ and $C = \partial D$, and let $R$ be the compact annulus in $A$ with $\partial R = C' \cup C$.
Question: What is the image of $R$ under the quotient map $p$?
Answer: That image $p(R)$ is homeomorphic to the quotient space of $R$ where opposite points of $C$ are identified, and one can show by direct construction that it is homeomorphic to a Möbius band which I'll denote $M = p(R)$. Thus, removing $D$ and gluing opposite boundary points of $C$ results in the same surface as removing $D'$ and gluing in the Möbius band $M = p(A)$ by identifying the circle $p(C') \subset p(A)$ with the boundary of $M$ by a homeomorphism.
For (b) the formulation is not quite correct: one needs to be careful with orientations. Let me assume that $A$ is orientable, in which case one needs to choose an orientation on $A$ which induces an orientation on $A-(D_1 \cup D_2)$ and thereby induces boundary orientations on each circle $C_i = \partial D_i$, and then one needs to require that the gluing homeomorphism $C_1 \mapsto C_2$ reverses boundary orientations. The danger is that if you instead preserve boundary orientations then the result is $A \# K$, where $K$ is the Klein bottle. (The case that $A$ is nonorientable is more complicated, leading to the conclusion that $A\#T$ and $A\#K$ are homeomorphic).
So, with that proviso about orientations, there is a similar answer but one needs a preliminary construction. Let $D_1,D_2$ be the two open discs that are removed, and denote their boundary circles $C_i = \partial D_i$. What you need next is that there exists an open disc $D \subset A$ which "engulfs" both of $D_1,D_2$, in the sense that in an appropriate coordinate system on $D$, the discs $D_1,D_2$ are disjoint round subdiscs of $D$. One way to see this, using path connectivity, is to construct an embedded path $\gamma$ in $A$ whose intersection with $D_i$ is a single point of $C_i$; one can then take $D$ to be a regular neighborhood of $D_1 \cup \gamma \cup D_2$, which one can show by direct construction is homeomorphic to a disc with subdiscs as said.
Once you have $D$, now let $P$ be the compact "pants" subsurface of $D$ with $\partial P = \partial D \cup \partial D_1 \cup \partial D_2$. One can show, again by direct construction, that when the quotient map $p : A - (D_1 \cup D_2) \to F$ is restricted to $P$, the image $p(P)$ is a torus.