Is this the right way to construct the adjunction space homotopy equivalent to $S^1$? If not, then how to do it? Currently, I'm working on an example from a topology book which states that The sphere $S^n$ can be obtained by attaching an $n$-cell to a space with one point: $D^n\cup_f\{a\}$.

Question:
I want to start with $Y=S_1\cup S_2$ s.t. $S_1,S_2\cong S^1$ (i.e. $S_1,S_2$ are homeomorphic with $\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2=1\}$) and $S^1\cap S^2=\{p\}$. Then I want to construct a space $Y\cup_f D^2$ homotopy equivalent to $S^1$.

Choose a subset $K\subset Y$ and from that, define a continuous map and use the equivalence relation generated by that relation to glue the two spaces together.
How do I choose that $K$? And which $f$ should be used to define the equivalence relation?
(I choose $S^1$ to be the goal because it is a relatively simple situation for this problem.)

My Attempt:
First I have $S_1,S_2\cong S^1$ (i.e. $S_1,S_2$ are homeomorphic with $\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2=1\}$) and $S^1\cap S^2=\{p\}$. I denote this space $Y=S_1\cup S_2$, and then I need to construct $Y\cup_f D^2\simeq S^1$.
I tried different kind of closed subset $K$ of $Y$:


*

*$K$ containing both part of $S_1$ and $S_2$ s.t. $K\cap S_1\supseteq\{p\}$ and $f:K\to \partial D^2$ to be a continuous mapping such that $f(K)$ is an arc on $\partial D^2$, I tried to identify those point by the equivalence relation $\sim$ generated by $ k\sim f(k),\forall k\in K$. And the graph is like the picture below and is obviously not homotopic equivalent to $S^1$.






*Select $K=S_2$ and then the Urysohn Lemma ensures that $\exists f:K\to\partial D^2$ which is a continuous mapping to an arc of $D^2$, to be specific, donote that arc to be $A$ and $f$ maps $K$ onto the arc $A$. Then, I got the following diagram by the universal property of the adjunction space. Identify the equivalence class $\simeq$ generated by $k\simeq f(k)\simeq f^{-1}(\{f(k)\})$

The resulting space is a small circle ($S_1$) and $D^2$. Define a retraction on $Y\cup_\phi D^2 $, so that we can retract $D^2$ to the connecting point $p$, which means $Y\cup_\phi D^2\simeq S^1$. But this seems invalid. since there are two layers by the disjoint union, then I don't know how to visualize the result because it seems to be a very strange space......

My Problem:
I'm not sure if this construction is valid, because $Y$ and $D^2$ are both subset of $\mathbb{R}^2$ and I wonder in this case
1) How to choose the right $K\subset Y$?
2) How to visualize and test if the resulting space is what I've been looking for?
Note: I'm pretty sure that there exists such $K$ since there is an exercise question asking for this.
Any help will be appreciated. Thanks in advance for your time and effort!
 A: I don't understand what you are trying to do with the Urysohn lemma, but it seems to me that the method is very simple. 
Choose one of your two circles, say $S_2$. Since it is a circle, it is homeomorphic to the border of a disk $D^2$. Just glue the disk on the circle. It makes the $S_2$ part of $Y$ contractible, so the resulting space is homotopy equivalent to $S_1=S^1$.
A: Your second construction is indeed valid. 
There is a convenient term for your space $Y$, it is called the "wedge product" of two copies of $S^1$, denoted by $S^1 \vee S^1$. In general, for two pointed spaces $(A,a_0), (B,b_0)$ we define the wedge product as
$$ A\vee B = A\sqcup B /\{a_0, b_0\}. $$
That is, take their disjoint union, and attach them together at their basepoints, which for us is some arbitrary $p\in S^1$. A very nice property of this construction is that if $B$ is contractible then $A\vee B$ is homotopy equivalent to $A$ (proof: exercise).
Now for your space $Y = S^1 \vee S^1$, if you choose $K$ to be the subset $\{p \} \vee S^1$ and attach a disc $D^2$ to $K$ by mapping its boundary identically onto the second copy of $S^1$ the resulting space is $$Y' = Y \cup_{id_2} D^2 \cong S^1 \vee D^2 \simeq S^1$$
(where the basepoint of $D^2$ is also $p$).
