Piece-wise smooth Rectifiable Jordan Curve with Grid Overlay. I'm looking at the proof of Cauchy's theorem $\oint_{C}f(z)dz = 0$ in Titchmarsh's Theory of Functions (second edition p.75). This assumes $f$ is holomorphic and $C$ is a piece-wise smooth rectifiable closed simple curve.    
He puts a grid over the curve and its interior to divide it into "a large number of small parts" and then separately shows that the integral round each of the interior squares is zero and that the sum of the integrals round the squares intersecting the curve is zero.
The proof seems to assume that for squares smaller than some size  the curve will always intersect each square with a single connected arc in order to justify forming a closed contour with with the sides of the square interior to the curve. Is this assumption true and if so how is it proved ? If true does it still hold without the smoothness assumption ? 
I have tried several approaches to proving this without any success. Considering $C$ as a rectifiable Jordan curve didn't get me anywhere, nor did knowing that disjoint compact sets have a minimum separation. 
 A: The older books rely on intuitive geometric arguments to prove Cauchy's theorem for an arbitrary Jordan curve.  Despite the importance of the theorem, the authors prefer to avoid the lengthy discourse.
Copson even goes so far to state that the final step in proving the theorem for a general (as opposed to polygonal) region is " ... one of the most difficult things to prove in the whole of the theory of functions of a complex variable," and then simply moves on.
Nevertheless, the general theorem can be proved without invoking any deep topology in a series of steps for (1) triangles, (2) convex polygons, (3) closed polygonal Jordan curves, (4) arbitrary closed polygonal curves, and, finally, (5) simple close rectifiable curves (Jordan curves).
The first four steps are straightforward-- with the latter steps somewhat more involved with respect to geometric arguments.
Nevertheless, the supposedly difficult final step is facillitated with the following lemma:

Suppose $f:D \subset \mathbb{C} \to \mathbb{C}$ is continuous  and $C
 = \{\phi(t): a \leqslant t \leqslant b\}$ is any rectifiable curve in $D$ defined by a continuous function $\phi:[a,b] \to \mathbb{C}$. 
  Given $\epsilon > 0$, there exists $\delta = \delta(\epsilon)> 0$ such
  that for any partition $P= \{a< t_0 < t_1 < \ldots < t_n = b\}$ with
  $\|P\| < \delta$, the polygonal curve $\Gamma
 =\overline{\phi(t_0)\phi t_1)} + \ldots + \overline{\phi(t_{n-1})\phi(t_n)} $  is contained in $D$ and
$$\left|\int_C f(z) \, dz - \int_\Gamma f(z) \, dz \right| <
 \epsilon.$$

Having proved Cauchy's theorem for analytic functions and polygonal curves (such as $\Gamma$) it easily follows using the lemma that the theorem is true for arbitrary Jordan curves $C$.
The proof of the lemma is elementary and sketched below.
(1) Using compactness of $C$ find a compact set $E \subset D$ and $\rho > 0$ such that the closed disks $\bar{D}(z;\rho) \subset E$ for all $z \in C$.
(2) Using uniform continuity of $f$ on $E$ find $\eta > 0$ such that $|f(z) - f(z')| < \epsilon/(2L)$ when $|z - z'| < \eta$ and where $L$ is the length of $C$.
(3) Using uniform continuity of $\phi$ on $[a,b]$, find $\delta_2 > 0$ such that $|\phi(t) - \phi(t')| < \min(\rho,\eta)$ when $|t-t'| < \delta_2$.
(4) There exists $\delta_1 > 0$ such that for any partition $P = (a= t_0 < t_1 < \ldots < t_n = b)$ with $\|P\| < \delta_1$, we have (since $f$ is integrable),
$$\tag{*}\left| \int_C f(z) \, dz - \sum_{k=1}^n f(\phi(t_k))(\phi(t_k) - \phi(t_{k-1}))\right| < \frac{\epsilon}{2}$$
(5) Finally, by requiring further that $\|P\| \leqslant \min(\delta_1,\delta_2)$, it can be shown using (2) and (3) that,for the polygonal curve $\Gamma
 =\overline{\phi(t_0)\phi t_1)} + \ldots + \overline{\phi(t_{n-1})\phi(t_n)} $,
$$\tag{**}\left| \int_\Gamma f(z) \, dz - \sum_{k=1}^n f(\phi(t_k))(\phi(t_k) - \phi(t_{k-1}))\right| < \frac{\epsilon}{2}$$
In view of (*) and (**), the lemma is proved.  
