Suppose $X$ is path-connected, then for any $y \in X$, and for any $x$, there is a path $\phi_x : I \to X$ such that $\phi_x(0)=x$ and $\phi_x(1) = y$.
With those maps, you can form a map $\phi : X \times I \to X$, with $\phi(x,t) = \phi_x(t)$.
This map has all the properties you would want of a deformatino retract from $X$ onto $y$, except that it has no reason at all to be continuous.
An example of a non retractable path connected space is the circle.
Let's say you pick $y = (1,0)$, and decide to connect any $x$ to $y$ in a clockwise motion, via the top of the circle.
Then the map you $\phi$ you will obtain will not be continuous at $(y,t)$ for any $t \in ]0;1[$, since the neighboor points below $y$ will travel all around the circle, while $y$ stays where he is.
And as Hatcher says, it is not trivial to show that the circle is not retractable onto a point.
Here is an idea for how to make this precise.
Consider the composition $X\times I\to X\to X/R$ of $H$ with the quotient map $q$:
\begin{matrix}
X\times I & \stackrel{H}\to & X \\
\downarrow & &\downarrow \\
(X/R)\times I & \stackrel{G}\to & X/R
\end{matrix}
The deformation retract $H$ induces a deformation retract $G$ since by definition $H$ must satisfy $H(a,t)=a$ for every $a$ on the border, and every $t\in I.$ To see this, we can consider an induced relation on $X\times I$ defined by $(x,t)\sim(y,s)$ iff $t=s$ and $xRy$, and then use the fact that $q\circ H$ is constant on equivalence classes of ${X\times I\over\sim}\cong(X/R)\times I$ (this follows from the fact that $I$ is locally compact; more information is provided in this StackExchange) to see that it factors through the quotient $q':X\times I\to (X/R)\times I$. Proving $G$ is actually a deformation retract from here should be trivial.
Best Answer
One general way to prove nonexistence of maps is to assume the map exists, apply one of your favorite algebraic topology functors, and thereby obtain a contradiction. That's how the proof of the "no retraction" theorem goes for nonexistence of a retraction from a closed 2-dimensional disc to its boundary circle, a proof which you probably know (and which is a key step in proving the Brouwer fixed point theorem).
In this situation, suppose that $M$ is a Möbius band with boundary circle $\partial M$. To say that a function $f : M \mapsto \partial M$ is a retraction means that the composition $$\partial M \xrightarrow{i} M \xrightarrow{f} \partial M $$ is equal to the identity map, where $i$ is the inclusion. Now let's apply our favorite functor, the fundamental group functor. It follows that the composition $$\underbrace{\pi_1(\partial M)}_{\mathbb Z} \xrightarrow{i_*} \underbrace{\pi_1(M)}_{\mathbb Z} \xrightarrow{f_*} \underbrace{\pi_1(\partial M)}_{\mathbb Z} $$ is equal to the identity map. What could possibly go wrong?
Well, we know that $i_*$ is the "times $2$" homomorphism of $\mathbb Z$, because the fundamental group of $M$ is generated by the curve that goes around the core of $M$, and the boundary of $M$ is homotopic to the curve that goes $2$ times around the core.
Now if you work through the algebra, you will discover that no matter what homomorphism of $\mathbb Z$ you compose after the "times $2$" homomorphism, you will never get the identity homomorphism of $\mathbb Z$. That's the contradiction which proves that you cannot retract $M$ to $\partial M$.
As you study algebraic topology more deeply, you will see this pattern repeated over and over: new algebraic topology functors put to use in disproving existence of maps of various sorts.