Topological dimension of topologist's sine curve and general question about topological dimension I am working on Exercise 3 in section 50 of Munkres' Topology. The exercise is to show that the topologist's sine curve ($S=\{(x,sin(\frac{1}{x}):0<x\leq1\}\cup\{0\}\times[-1,1]$) has topological dimension 1.
Let $\mathcal{A}$ be an open cover of $S$
Since $S$ is a closed and bounded subset of $\mathbb{R}^2$ it is compact, hence we should be able to use the Lebesgue number lemma to form small subsets of the sets in $\mathcal{A}$ (Hence forming a refinement $\mathcal{B}$ of $\mathcal{A}.$ I am having trouble constructing a refinement in a clean way such that no more than 2 sets intersect however. What I have so far is:
By the Lebesgue Number Lemma, there exists
$\delta>0$ such that any open subset $U\subset S$ with diameter
less than $\delta$ will be an element of some $A\in\mathcal{A}$.
Now choose $n$ so large that $\frac{1}{n}<\delta$ and cover the
vertical line $\{(0,y):-1\leq y\leq1\}$ with open squares of width
$\frac{1}{n}$ by placing their centers at the points $\{(0,sin(2n)+k(\frac{1}{n}-\epsilon)):k\in\mathbb{Z},\ \epsilon\ll1\}$
so that the squares have very small intersection areas and the sine
curve only interesects a single square. Then cover the remainder of
the sine curve with squares whose diameters are so small that they
are less than $\delta$, and the square covering $(\frac{1}{2n},sin(2n))$
does not only intersects the single square centered at $(0,sin(2n))$
on the vertical line. Because the portion of the curve from $[2n,1]$
is homeomorphic to $[2n,1]$ it has dimension 1, so if any three or
more squares intersect on the sine portion of our curve, it can be
refined to ensure no more than two do. Therefore the order of this
refinement will be order 2, hence the dimension of the entire curve
is 1, as desired. 
Is this a good way to go with the problem? I am not sure it is clear that you can refine the remainder of the curve to the right of the vertical line arbitrarily. 
Is there a different, more general way I could approach this problem? I am interested if there is some connection between the topological dimension of a set and its closure, but I could not find any info on this. For instance if it were the case that the topological dimension does not change after taking the topological closure under reasonable regularity of the topology that could be quite useful.
 A: 
cover the vertical line $\{(0,y):-1\leq y\leq1\}$ with open squares of width $\frac{1}{n}$ by placing their centers at the points $\{(0,sin(2n)+k(\frac{1}{n}-\epsilon)):k\in\mathbb{Z},\ \epsilon\ll1\}$ so that ... the sine curve only interesects a single square.

I do not understand what means “the sine curve only interesects a single square”.  If an open square covers a point of the vertical line then it is intersected by the sine curve.  

Is this a good way to go with the problem? I am not sure it is clear that you can refine the remainder of the curve to the right of the vertical line arbitrarily.

I do not understand all details of your construction, but it seems that a refinement problem can occur if there are points on the sine curve $(x,\sin\frac 1x)$ not covered by the refined cover of the vertical line, but belonging to a closure of two squares of this cover. 

I am interested if there is some connection between the topological dimension of a set and its closure, but I could not find any info on this. For instance if it were the case that the topological dimension does not change after taking the topological closure under reasonable regularity of the topology that could be quite useful.

Conversely, the topological dimension essentially depends on the closedness. For instance, each Tychonoff space $X$ with $d(X)<\frak c$ contains zero-dimensional dense subspace. On the other hand, the sum theorem [E, 1.5.3] states that if a  separable metric  space $X$  can  be represented  as  the  union  of  a sequence $F_l, F_2,\dots, $ of  closed subspaces  such  that  $\operatorname{ind} F_i\le n$ for  i = 1,2, ... then  $\operatorname{ind} X \le n$. 
You can also use the last theorem to show that topological dimension of topologist's sine curve is $1$.
I recall that “It  is possible  to  define  the  dimension of a topological space $X$  in  three different  ways,  the  small inductive dimension $\operatorname{ind} X$,  the  large  inductive dimension  $\operatorname{Ind} X$, and the  covering dimension  $\operatorname{dim} X$. The three  dimension functions coincide in the class of  separable metric spaces, i.e.,  $\operatorname{ind} X = \operatorname{Ind} X =\operatorname{dim} X$ for  every  separable metric  space $X$.  In  larger  classes  of  spaces the  dimensions  $\operatorname{ind}$, $\operatorname{Ind}$, and  $\operatorname{dim}$ diverge”.  [E, p. V]
References 
[E] Ryszard Engelking Dimension Theory, North-Holland mathematical library 19, Elsevier, Academic Press, 1978. (available at LibGen).
