The circle is not contractible I know that the circle is not contractible because I know that $\pi_1(S^1)\cong \mathbb Z$. But something is going wrong in my head. Choose a basepoint $*$ on the circle and chose an orientation (clockwise say) and for each point on the circle we take the path from that point to the base point $*$ going clockwise and making only one tour, meaning the first time we meet $*$ we stop. This seems to give us a continuous way to contract the circle onto $*$, making the circle contractible,  what is wrong with this reasoning ?
 A: It seems to me that in the usual polar coordinates on the circle, OP is talking about the homotopy
$$
H: S^1 \times  [0, 1] \to S^1 : (\theta, s) \mapsto (1-s) \theta.
$$
For $s > 0$, this map is not continuous at $\theta = 0$, because no neighborhood of $0$ maps to a small neighborhood of $H(0, s)$. Hence it's not a deformation retraction from $S^1$ to a single point. 
Post-comment addition: 
Let's look at $s = \frac{1}{2}$. Fixing $s =\frac{1}{2}$, our map looks like
$$
\theta \mapsto H(\theta,\frac{1}{2}) = \frac{1}{2} \theta.
$$
Let's therefore give it a name, and say
$$
f(\theta) = \frac{1}{2}\theta.
$$
Pick a small neighborhood of $0$, say, 
$$
U = (0 \le \theta < 0.1) \cup (2\pi - 0.1 < \theta < 2\pi)
$$
Under the map $f$, $0$ goes to $0$ (i.e., $f(0) = 0$). So the preimage of $U$, under $f$, i.e., $f^{-1}(U)$ should contain a neighborhood of $0$. 
OK: small admission here. I should really be saying "for any neighborhood $U$ of $0$, there's a neighborhood $V$ of $0$ with the property that $f(V) \subset U$." Instead, I'm going to look at $f^{-1}(U)$ and note that it "looks wrong", i.e., that if doesn't contain any neighborhood $V$ of $0$.
What is $f^{-1}(U)$? Well, $U$ is a union of two parts. The preimage of the first part is 
$$
\{ \theta \mid 0 \le \theta < 0.2 \}.
$$
Clear enough, I hope. 
The preimage of the second part is empty. 
That means that the preimage of $U$ contains no points that are just below $2\pi$, hence it cannot contain any neighborhood of $\theta = 0$, since all neighborhoods of $\theta = 0$ contain points with $\theta$ arbitrarily near $2\pi$. 
A: You have a family $(\gamma_t)_{t\in[0,1]}$ of continuous maps from the circle to itself such that $\gamma_0$ is the identity, $\gamma_1$ is constant, and $\gamma_t\to\gamma_1$ pointwise is $t\to1$. But contractibility requires that $\gamma_t\to\gamma_1$ uniformly.
At least that's the way I read your description. It's sufficiently vague that it could be the family of maps you have in mind is not the one I thought you meant.
To clarify the ambiguity: Let's talk about the interval $[0,1]$ with the points $0$ and $1$ identified.
It could be you mean something like $$f_t(x)=x^{1-t}.$$That satisfies $f_0(x)=x$, $f_1(x)=1=0$, and each $f_t$ is continuous. That's what I was assuming above; the problem is that $f_t\to f_1$ pointwise but not uniformly.
Or it could be you mean something like $$f_t(x)=x(1-t),$$in which case $f_t\to f_1$ uniformly but $f_t$ is not continuous. (More accurately, $f_t$ is not even well-defined on our quotient space, since $f_t(0)\ne f_t(1)$. Defining a similar map on $[0,1)$ give something discontinuous, if we note that points close to $1$ are close to $0$.)
A: As you try this on points farther and farther along the way, you start approaching the point $*$ from the other side. To be continuous also there, you'd have to drag $*$ itself around the circle once, and then the points slightly past it once and a bit. But that's in contradiction to your rule that you only move them until you hit $*$ the first time. So your contraction will rip the circle right at the point $*$, violating continuity.
A: it is not continuous at $*$. You construct an homotopy $H_t:S^1\rightarrow S^1$ such that $H_t(x)=c_t(x)$ where $c_t$ is the path going towards $*$ from $x$ clockwise. You must have $H_t(*)=*$. There exists a sequence $lim_nx_n=*$ anticlockwise such that $x_n\neq *$ and $lim_nH_t(x_n)=c_t(x_n)=*$ anticlockwise otherwise $H_t(S^1), t<1$ is homeomorphic to a non  closed interval. Impossible since the image of $S^1$ by $H_t$ has to be compact. Thus $H_t(S^1)=S^1, t<1$, this implies that $H_1(S^1)=S^1$ To see this let $x\in S^1$, and $t_n$ such that $lim_nt_n=1,t_n<1$ there $x_n$ such that $H_{t_n}(x_n)=x$. You can find a subsequence $x_{i(n)}$ of $x_n$ which converges towards $y$, $H_1(y)=lim_nH_{t_{i(n)}}(x_{i(n)})=x$. Contradiction since $H_1(S^1)=*$.
