You can go both ways with quotients here: $X$ (the unit circle) can be seen as a quotient image of $Y$ (a closed interval), by identifying end points. I suppose you covered this already.
You can also go the other way: restrict the projection onto the x-axis to $X$, and then the image is $[-1,1]$. This is a quotient map because the projection is continuous and the domain $X$ is compact.
Another way to see this, is to define an equivalence relation $R$ on $X$ by declaring all pairs $(x,y)$ and $(x, -y)$ to be equivalent, and no others. Then $X/R \simeq Y$. This is essentially the same idea.
To expand on this last idea: we have defined an equivalence relation by specifying all classes (all of them doubletons, except that $(-1,0)$ and $(1,0)$ are in a class by themselves (as $-0 = 0$). The set of all these classes is called $X/R$ (so it is a set of sets).
There is a standard map $q$ from $X$ to $X/R$, defined by sending a point $x$ to its class $[x]$ under $R$. E.g. $q((0,1)) = \{(0,1), (0,-1)\}$ (this is a point in $X/R$, recall!), $q((1,0)) = \{(1,0)\}$. The topology on $X/R$ (the quotient topology) is defined by $O \subset X/R$ is open iff $q^{-1}[O]$ is open in $X$. As $X$ already has a topology (inherited from being a subset of the plane), this defines a topology on $X/R$ (needs a small proof, but I assume this is in your text). By definition $q$ is then also a continuous map, so this makes $X/R$ a continuous image of $X$.
This also has the nice property that to see that a function $f: X/R \rightarrow Y$ for some space $Y$ is continuous for this topology, it suffices to check that $f \circ q$ is continuous between $X$ and $Y$: if $O$ is open in $Y$, $f^{-1}[O]$ should be open in $X/R$, but this just means by definition that $q^{-1}[f^{-1}[O]]$ is open in $X$, but the latter is exactly $(f \circ q)^{-1}[O]$, so if $f \circ q$ is continuous, we know that $f^{-1}[O]$ is indeed open.
Now we define a function f on $X/R$ to $[-1,1]$ as follows: for a class $[(x,y)]$ define $f([(x,y)]) = x$ (so we take the projection onto the first coordinate, really). We check first this is actually well defined: if a class in $X/R$ has more than one element, it in fact has two , and they are of the form $(x,y)$ and $(x,-y)$ (this is how we defined $R$). Whichever one we choose as the representative for the class (as $[(x,y)] = \{(x,y), (x,-y)\} = [(x,-y)]$), the $x$-coordinate is the same, so we get the same value in $[-1,1]$ in either case.
And the function $f \circ q$ is just the normal function $(x,y) \rightarrow x$, which we know to be continuous, so $f$ is continuous by the previous.
Also, $f$ is 1-1: if $f([(x,y)]) = f([(x',y')]$ this means that $x = x'$. And two points on the unit circle with the same $x$-coordinate can only differ in sign on the $y$-coordinate (as $y = \pm \sqrt{1-x^2}$), and so these points are in the same class, hence are really the same class in $X/R$ (!).
And a 1-1 continuous function from a compact (as $X/R$ is the continuous image of the compact $X$) space to a Hausdorff space ($[-1,1]$) is a homeomorphism so $X/R$ is homeomorphic to $[-1,1]$ under $f$.
Best Answer
The key ingredient to use here is compactness. Suppose no such disk exists. That means that for each $\epsilon>0$, $\overline{B_{r+\epsilon}(z_0)}$ is not contained in $A$, so there exists $x\not\in A$ such that $d(x,z_0)\leq r+\epsilon$. Taking a sequence of such points as $\epsilon\to 0$, we get a sequence $(x_n)$ of points, none of which are in $A$, such that $d(x_n,z_0)$ converges to $r$. This is a bounded sequence in $\mathbb{C}$, so it has a subsequence which converges to some limit $x$. Then $d(x,z_0)=r$. Also, since each $x_n$ is not in $A$ and $A$ is open, $x\not\in A$. But now this is a contradiction since $\overline{B_r(z_0)}\subseteq A$.