The space obtained by attaching a subspace of a topological space to itself. Just asking a special case of attaching, it seems pretty vague to me.
Let $Y$ be a space and $X$ be its subspace. Then the space obtained by $Y$ attaching its subspace $X$ via $\iota$ (the embedding) is defined to be the quotient of the coproduct $X\coprod Y/\sim $ in which $\sim$ is the equivalence relation generated by the relation $x \sim \iota(x)$. Can u tell me what this space is explicitly?
Intuitively, I think the answer is just $Y$, I imagine $X\coprod Y$ is just the disjoint union of two sets, and the equivalence relation implies that every two identical points $(x,\iota(x)) \in X\times Y$  is the same. So to sum up, the result is Y.
 A: What are you actually asking about is called the adjunction space.

Definition: (Adjunction Space) Suppose $A$ and $B$ are topological spaces and $W$ is a closed subspace of $B$ and $f : W \to A$ is a continuous map. Let $\sim$ be the equivalence relation on the disjoint union $A \sqcup B$ generated by $(w, 2) \sim (f(w), 1)$ for all $w \in W$ and denote the resulting quotient space by $A \cup_f B = (A \sqcup B) / \sim.$ Any such quotient space is called an adjunction space and is said to be formed by attaching $B$ to $A$ along $f$.

Now specializing to your question above we have the inclusion map $i : X \to Y$ defined by $i(x) = x$ and note that since $X$ is a closed subspace of itself and $i$ is continuous we can form the adjunction space $$Y \cup_i X = \left(Y \sqcup X\right) / \sim$$ where $\sim$ is the relation generated by $(x, 2) \sim (i(x), 1)$ for $x \in X$. But then by definition of $i$ it follows that $\sim$ is the relation generated by $(x, 2) \sim (x, 1)$ for $x \in X$. 
Before we can quotient out the disjoint union, we first need to see what this relation actually is. The equivalence relation generated by $(x, 2) \sim (x, 1)$ for $x \in X$ is the smallest equivalence relation on $Y \sqcup X$ that has this property. So basically under this equivalence relation for a fixed $x \in X \subseteq Y$ it follows that $(x, 2)$ and $(x, 1)$ belong to the same equivalence class in $\left(Y \sqcup X\right) / \sim$ and each $(y, 1)$ has it's own equivalence class $\{(y, 1)\}$ for all $y \in Y \setminus X$. So set-theoretically we have that 
\begin{align*}\left(Y \sqcup X\right) / \sim \ &= \left\{[ (x, 1)] \ | \ x \in X \subseteq Y \right\} \cup \left\{[(y, 1)] \ | \ y \in Y \setminus X\right\} \\
&= \left\{\{(x, 2), (x, 1)\} \ | \ x \in X \subseteq Y \right\} \cup \left\{\{(y, 1)\} \ | \ y \in Y \setminus X\right\}
\end{align*}
We now appeal to the following theorem to finish off this proof.

Theorem: (Uniqueness of quotient spaces) Suppose $q_1 : X \to Y_1$ and $q_2 : X \to Y_2$ are quotient maps that make the same identifications (i.e $q_1(x) = q_1(x') \iff q_2(x) = q_2(x')$). Then there is a unique homeomorphism $\varphi : Y_1 \to Y_2$ such that $\varphi \circ q_1 = q_2$

Let $q : Y \sqcup X  \to Y \cup_i X$ be the natural quotient map defined by $q(a, i) = [(a, i)]$. 
Appealing to the uniqueness of quotient spaces this motivates us to define a quotient map $f$ such that $f(x,2) = f(x,1)$ for all $x \in X$ and $f(y, 1) = (y, 1)$ for all $y \in Y \setminus X$. The projection map $f : Y \sqcup X \to Y$ defined by $f(a, i) = a$ is precisely the map that achieves this outcome. Note that $f$ is quotient map, the fact that $f$ is a quotient map follows from the definition of the disjoint union topology.
We claim that $f$ and $q$ make the same identifications. To prove this we need to show that $q(a, i) = q(b, j) \iff f(a, i) = f(b, j)$. I will prove the forward direction, the reverse direction follows similarly.
Suppose that $q(a, i) = q(b, j)$. 
If $(a, i) = (x, 2)$ for some $x \in X$ then $(b, j) = (x, 2)$ or $(b, j) = (x, 1)$ (by definition of the relation generated by $\sim$ and $q$). In either case $f(a, i) = f(x, 2) = x$ and $f(b, j) = x$, hence $f(a, i) = f(b, j)$ as desired.
If $(a, i) = (x, 1)$ for any $x \in X$, then $(b, j) = (x, 1)$ or $(b, j) = (x, 2)$ (again by definition of the relation generated by $\sim$ and $q$). In either case we have $f(a, i) = f(x, 1) = x = f(b, j)$ as desired.
If $(a, i) = (y, 1)$ for any $y \in Y$, then $(b, j) = (y, 1)$ (again by definition of the relation generated by $\sim$ and $q$). Hence $f(a, j) = f(y, 1) = y = f(b, j)$ as desired. 
Since these cases exhaust all possible choices for $(a, i)$ and $(b, j)$ for which $q(a, i) = q(b, j)$ it follows that $q(a, i) = q(b, j) \implies f(a, i) = f(b, j)$ proving the forward direction.
After proving the reverse direction it follows that $q$ and $f$ make the same identifications and since they are both quotient maps, by the theorem above it follows that the following diagram commutes

and that there exists a unique homeomophism $\varphi$ from $Y \cup_i X$ to $Y$ hence $$Y \cup_i X \cong Y.$$ 
