Hatcher's Algebraic Topology 0.6(a) 
Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the horizontal segment
  $[0, 1] \times \{0\}$ together with all the vertical segments $\{r\}\times[0, 1 - r]$ for
  $r$ a rational number in [0, 1]. Show that $X$ deformation retracts to
  any point in the segment $[0, 1]\times \{0\}$, but not to any other point. [See
  the preceding problem.]

Proof: Consider the homotopy $ \forall p \in X$,
$$f_t(x,y) = (1-t)(x,y) + (x,(1-t)y).$$
Hence we find the family of maps $f_t: X \to X, t \in I$, such that $f_0 = \mathbb{I}$ (the identity map), $f_1(X) = [0, 1]\times \{0\}$, and $f_t|[0, 1]\times \{0\}=  \mathbb{I}$ for all $t$.
Now we show not to any other point.
First consider $q_1 = (x,y)$ where $y > 1 - x$. Then this is an isolated point, so it can not be retracted.
Then consider $q_2 = (x,y)$ where $ x \in \mathbb{R - Q}$. How can I show this won't retract?
Also, I am concerned that my proof so far does not use the preceding problem, that

If a space $X$ deformation retracts to a point $x ∈ X$, then for each neighborhood $U$ of $x$ in $X$ $\exists$ a neighborhood $V ⊂ U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.

Thank you very much for your help!

Your homotopy for the first part is not quite correct. (Note that $f_0(x, y)=(2x, 2y)$). For your other argument $q_1, q_2$ are not in the space so you don't need to consider them. The question is asking you to show that it doesn't retract onto any other point $\textit{in the space}$. 
So if you pick another point that is not on the $x$-axis you can choose a sufficiently small neighborhood such that it is not path-connected. This should do it for you, using the previous property about inclusions being nullhomotopic.
