Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$. 
Suppose that $X$ is homotopy-equivalent to a one-point space, then each point of $X$ is a deformation retract of $X$.

This was my (failed) attempt to solve this problem.
Suppose that $X \simeq \{c\}$. Pick a point $p \in X$. We want to show that there exists a retract $r : X \to\{p\}$ such that for the map $i_{\{p\}} : \{p\} \to X$ defined by $i_{\{p\}} (p) = p$ we have $i_{\{p\}}  \circ r \simeq \text{Id}_X$.
Let $r : X \to \{p\}$ be defined by $r(x) = p$ for all $x \in X$. We claim that this is the required retraction. Then $i_{\{p\}} \circ r =r$. So the problem boils down to showing that $r \simeq \text{Id}_X$.
Now since $X \simeq \{c\}$, there exist continuous maps $f : X \to \{c\}$ and $g : \{c\} \to X$ such that $f \circ g \simeq \text{Id}_{\{c\}}$ and $g \circ f \simeq \text{Id}_{X}$. Now note that there can only be one such map $f$, the map defined by $f(x) = c$ for all $x \in X$ (which is trivially continuous).
At this point I figured that if I showed that $r \simeq g \circ f$, then I'd arrive at $r \simeq \text{Id}_X$. But for any $x \in X$ we have $(g\circ f) (x) = g(f(x)) = g(c)$ and since we don't know exactly what function $g$ is, $g(c)$ could be anything.
Now I'd have to define an explicit homotopy betwen $r$ and $g \circ f$ but I'm not sure how to do that. Intuitively I need to find some way to continuously 'move' $g(c)$ to $p$ in $X$ via this homotopy, and usually if $X$ is convex then this can be done using a straight line homotopy. I'm not sure how to construct any analogous homotopy in this case.
How can I prove the above?
 A: Obvious facts:
(1) There is only one function from $X$ to any one-point space $P$ which is automatically continuous.
(2) All one-point spaces are homeomorphic.
From (2) we conclude
(3) $X$ is homotopy equivalent to any one-point space $P$.
Now take $P = \{ p \}$ with $p \in X$. By (1) there is a unique map $r : X \to P$ which is of course a retraction. By (3) $r$ must have a homotopy inverse $i : P \to X$. It is characaterized by $i \circ r \simeq Id_X$ ($r \circ i = Id_P$ is trivial). Let $H : X \times I \to X$ be a homotopy such that $H_0 = Id_X , H_1 = i \circ r$. This is precisely what is defined as a deformation retraction from $X$ to $P$.
Note that any map $j : P \to X$ (which is given by $j(p) \in X$) is a homotopy inverse for $r$. This comes from the fact that $j \circ r = Id_X \circ j \circ r \simeq i \circ r \circ j \circ r = i \circ r \simeq Id_X$. 
As Peter Franek mentioned, $r$ is in general not a strong deformation retraction which would require that $H(x,t) = x$ for all $(x,t) \in P \times I$. In fact, a strong deformation retraction does not always exist.
A: 
There is a little fact that you missed. That is any space $X$ that homotopy-equivalence to any one point space is path-connected.

Suppose $X \simeq \{*\}$ . Let $f : X \to \{*\}$ be the homotopy equivalence and $g : \{*\} \to X$, defined as $g(*) = p$ for some $p \in X$, be the homotopy inverse for $f$. By hypothesis, the composition $h=g \circ f : X \to X$ is a constant map homotopic to $\text{Id}_X$. Let $H  : X \times I \to X$ be the homotopy, with $H(x,0) = \text{Id}_X(x)=x$ and $H(x,1) = h(x)=p$. Note that for a fix $x \in X$, $H(x,\cdot) : I \to X$ is a path from $x$ to $p$. So $X$ is path-connected.
Choose any $q \in X$, we want to show that $q$ is a deformation retract of $X$. That is we want to show that $\iota_q \circ r \simeq \text{Id}_X$, where $r : X \to \{q\}$ is a constant map (retraction) and $\iota_q : \{q\} \to X$ is the inclusion map. Let $k := \iota_q \circ r$.

As you say, we need to move the point $g(*) = p$ to $q$ continously. Luckly, $X$ is path-connected.

Note that $h = g \circ f$ and $k = \iota_q \circ r$ are both constant maps to point $p$ and $q$ (resp.). Since $X$ is path-connected, we have a path $\alpha : I \to X$ from $\alpha(0) = p$ to $\alpha(1) =q$. Define a map $F : X \times I \to X$ by $F(x,t) = \alpha(t)$. This means that $F(x,0) = \alpha(0) = p = h(x)$, and $F(x,1) = \alpha(1) = q = k(x)$ for any $x \in X$. So $h \simeq \iota_q \circ r$, and $h \simeq \text{Id}_X$ by hypothesis. Therefore $\iota_q \circ r \simeq \text{Id}_X$.
