Proving a criterion for connectedness The English translation of Markushevich's Theory of functions of a complex variable contains the following sufficient condition for the connectedness of a compact set on the complex plane.

Theorem 4.1 A nonempty compact set F is connected if given any two points $z,z'$
  in F and any $\epsilon >0$ there exists a finite number of points
  $z=z_0,z_1,\ldots,z_n=z'$ in F such that $|z_{k-1}-z_k| < \epsilon$ for $k=1,\ldots,n .$

I am interested in the converse of the above theorem, whose validity is assumed in Theorem 4.17 of the book but not proved.

Given a nonempty connected compact set F, and any two points $z,z'$ in it and any $\epsilon > 0$ there exists a finite number of points
  $z=z_0,z_1,\ldots,z_n=z'$ in F such that $|z_{k-1}-z_k| < \epsilon$ for $k=1,\ldots,n.$

I think I have a reasonably good plausibility argument for the correctness of the above, but I am struggling to convert to a proof.
My argument is:


*

*Choose $\delta > 0$ and consider all squares of width $\delta$ of the form $[k\delta,(k+1)\delta] \times [j\delta,(j+1)\delta]$ where k and j are integers.

*Let $\mathbb{U}$ be the collection of all squares above that intersect F, let the union of all sets in  $\mathbb{U}$ be U. 

*Then U is connected and its interior, G, is an open connected set that contains F.

*Since G is pathwise connected there is a continuous path $\gamma(t), 0 \leq t \leq 1$ in G with $\gamma(0) = z$ and $\gamma(1)=z'.$

*The uniform continuity of $\gamma$ implies the existence of points $z=z_0',z_1',z_2',z_{n-1}',z_n'=z'$ on the image of $\gamma$ such that $|z_{k-1}' - z_k'| \le \epsilon'$ for any given $\epsilon' > 0.$

*Replace each point in the finite sequence above (except the two
endpoints)  with a point in F that lies in a square in $\mathbb{U}$
containing it. Call the new sequence $z_0,\ldots,z_n$, by making
$\delta$ and $\epsilon'$ sufficiently small we can ensure
$|z_{k-1}-z_k| < \epsilon$ for $k=1,\ldots,n.$


Is the converse stated above corrected, if not, is there a nice counterexample?
If the converse is correct, is my argument above correct. I am suspicious about 3., I can "see" that its true for some simple cases, but I wonder if I am missing something.
Is there a simpler proof of the converse, if it is true?
Added Later:
From Eric's comments.


*

*Every point in F lies in the interior of U, since a point in F either lies in the interior of a square in $\mathbb{U}$ or lies in the corner or edge of a square in $\mathbb{U}$.  In the latter case the point lies in multiple squares in  $\mathbb{U}$ and is in the interior of their union.

*Any point in the interior of U can be connected with a straight line lying in the interior of U to a point in F, which implies the interior of U is an open connected set.


Rest follows as above.
 A: Your proof is correct.  To flesh out step 3 a bit, here is what you want to prove.  For every $x\in F$, you want to prove that a neighborhood around $x$ is contained in $U$.  For every $y\in G$, you want to prove that if $x\in F$ is in the same closed $\delta$-square as $y$, then there is a path from $x$ to $y$ which is contained in $G$.  Both of these are straightforward to prove, though they require a bit of casework if $x$ or $y$ is in the boundary of a square.  Given these facts, we can conclude that $G$ is an open set containing $F$, and every point of $G$ is in the same connected component of $G$ as some point of $F$.  Since $F$ is connected, this means $G$ is connected.  Since $G$ is open, this means $G$ is path-connected, so you can proceed to step 4.
You can get a similar and perhaps simpler proof along the lines you are suggesting by throwing out the $\delta$-mesh and instead just defining $G=\{z\in\mathbb{C}:|z-z'|<\delta\text{ for some }z'\in F\}$.  Then $G$ is open, contains $F$, and every point of $G$ can be path-connected to some point of $F$, so $G$ is connected and hence path-connected.  Now you can follow your proof starting from step 4.
There is another simple proof, which works for any connected metric space at all.  Suppose $F$ is a connected metric space and fix $\epsilon>0$.  Say that two points $z,z'\in F$ are $\epsilon$-connected to each other if there exists a finite number of points $z=z_0,z_1,\ldots,z_n=z'$ in F such that $d(z_{k-1},z_k) < \epsilon$ for $k=1,\ldots,n$.  It is easy to see that being $\epsilon$-connected is an equivalence relation on $F$.  Furthermore, if $d(z,z')<\epsilon$, then clearly $z'$ is $\epsilon$-connected to $z$.  This implies that every equivalence class is an open set.  But the complement of every equivalence class is also open, being the union of the other equivalence classes.  Thus since $F$ is connected, there can only be one equivalence class.  Thus any $z,z'\in F$ are $\epsilon$-connected.
