Let $A := \mathbb{S}^1 \subset \overline{\mathbb{B}^2}$, and let $f : A \hookrightarrow \overline{\mathbb{B}^2}$ be the inclusion map. Consider the adjunction space $\overline{\mathbb{B}^2} \cup_f \overline{\mathbb{B}^2}$ of $\overline{\mathbb{B}^2}$ to $\overline{\mathbb{B}^2}$ along $f$. I wish to prove that $\overline{\mathbb{B}^2} \cup_f \overline{\mathbb{B}^2}$ is homeomorphic to $\mathbb{S}^2$.
Notation: $\mathbb{S}^1$ is the unit circle, $\mathbb{S}^2$ is the unit sphere, $\overline{\mathbb{B}^2}$ is the closed unit ball; the adjunction space is defined as the quotient $ X \cup_f Y := \dfrac{X \sqcup Y}{a \sim f(a), \quad \forall a \in A \subseteq Y} $ for the attaching map $f : A \to X$.
My attempt: construct an explicit quotient map to $\mathbb{S}^2$ and conclude by uniqueness of quotients. Specifically, let $X := \overline{\mathbb{B}^2}$ and $Y := \overline{\mathbb{B}^2}$, where $A$ is now sitting in $Y$ and is embedded into $X$ via $f : A \hookrightarrow X$. I identify $X$ and $Y$ with actual closed balls $\{(x,y) \in \mathbb{R}^2 ~|~ x^2 + y^2 \leq 1\}$, and I define a map $ \varphi : X \sqcup Y \longrightarrow \mathbb{S}^2 $ $ (x,y) \in X \longmapsto \left(x,y, + \sqrt{1 - (x^2 + y^2)}\right), $ $ (x,y) \in Y \longmapsto \left(x,y, - \sqrt{1 - (x^2 + y^2)}\right). $ In other words, send one disc to the upper hemisphere, and the other to the lower hemisphere. Clearly $\varphi$ is surjective; it moreover makes the same identifications as $\sim$. It is therefore a quotient map, completing the proof by uniqueness of quotients.
Uniqueness of Quotients: If $\pi_1 : X \to Y_1$ and $\pi_2 : X \to Y_2$ are two quotient maps that make the same identifications (i.e. $\pi_1 (p) = \pi_1 (q) \iff \pi_2(p) = \pi_2(q)$), then there is a homeomorphism between $Y_1$ and $Y_2$.
Question 1: Are my deductions correct?
Question 2: Is there an easier way of doing this, i.e. without introducing explicit co-ordinates and maps?
Note: cannot use any homology theory, homotopy theory, etc..