A problem from Hatcher's algebraic topology This is Ex 1.1.19 from Hatcher's algebraic topology.

Show that if $X$ is a path-connected 1-dimensional CW complex with basepoint $x_0$ a 0-cell, then every loop in $X$ is homotopic to a loop consisting of a finite sequence of edges traversed monotonically. 

But consider the Hawaiian earring $X$. We can construct a loop $f:[0,1]\to X$, such that $f|_{[1/(n+1),1/n]}$ wraps around $n$th (largest) circle, and $f(0)=x_0$. $f$ is continuous. Isn't this a counter-exmaple to the problem?
 A: As the Lord Shark pointed out, the Hawaiian Earring is not a CW complex. If it had the weak topology it would be homotopic to the infinite wedge of circles, but the Hawaiian earring has the subspace topology induced by the Euclidean topology. This is exactly what makes the Hawaiian Earring special (not locally simply connected).
EDIT(on the weak topology): It may add to the confusion but to talk about weak topology we must talk about a space $X$ and maps $f:X \rightarrow Y$. Any subspace topology is the coarsest topology such that the inclusion map is continuous and is thus a weak topology with regards to that map. Thus it may be better to distinguish the Hawaiian Earring from a CW complex by noting it is not locally contractable, which a CW complex must be.
EDIT 2 ELECTRIC BOOGALOO 
To directly address your definition of weak topology: $A$ is open in $X$ iff $A \cap X^n$ is open. This should be tweaked, classically it is stated $ A$ is open iff $A \cap e_i^n$ is open for all $n$-cells $e_i^n$. As Dr. Mosher pointed out this is the strongest (read finest) topology such that the attaching maps are continuous.
Now let $A$ be such that $A \cap e_i^1$ is something like $[0,\frac{1}{2}) \cup (\frac{1}{2},1]$. Then $A$ is open but as $A$ contains the point $(0,0)$ it must contain infinite circles ( there must be an $j$ with $A\cap e_i^1 = e_i^1$ for all $i >j$ which is not the case so we arrive at a contradiction, the space does not have the weak topology ( in CW context.)
