Finite open covers of a complex $C^{(1)}$ curve. Consider a complex curve $\gamma \subset \mathbb{C}$, parametrized by $\alpha: [a,b]\to \mathbb{C}$, with $\alpha \in C^{(1)}$. Further, consider an finite open cover $\Phi$ of $\gamma=\alpha([a,b])$. Such cover exists by compactness and continuity. 
I want to show that we can find a $\textbf{finite}$ partition $a=t_0<t_1<\dots<t_n=b$ such that, for each $k\in\{1,\dots,n\}$, there exists $V\in \Phi$ with $\alpha \left ( [t_{k-1},t_k]\right ) \subset V$.
Intuitively, this seems obvious, but I can't seem to come up with a rigorous proof (or a counterexample). My best effort is as follows:
Since $\Phi$ is a cover of $\gamma$, $\alpha(a)$ belongs to some member of $\Phi$, call it $V_1$. If $\gamma \subset V_1$, then we are done. Suppose then that:
$$\{ t\in [a,b] : \alpha(t)\not \in V_1 \}\neq \emptyset$$
Since $\alpha$ is continuous and $V$ is open, the LHS is closed  (and bounded), hence compact, and thus contains a minimum element, say $t_1$, with $a<t_1$. Again, since $\Phi$ is a cover of $\gamma$, $\alpha(t_1)$ belongs to some member of $\Phi$ (different from $V_1$), say $V_2$. We repeat the argument to find $t_2 = \min\{ t\in [t_1,b] : \alpha(t)\not \in V_2 \}$, with $t_2 > t_1$. At each step, if the sets so far considered are already a cover for $\gamma$, we are done (alternatively, to avoid this we could suppose that $\Phi$ a minimal cover, which exists by finiteness). 
We can obviously continue this way to find $t_1<t_2<t_3,\dots$, but does the process end in a finite number of steps, where the final $t_n$ found is $b$? My objection to the past construction is that the relative lengths of the intervals $[t_{k-1},t_k]$ could become arbitrarily small, and hence we might never "reach" the endpoint $b$, ala Zeno's paradox. 
I imagine that I'm missing something related to the compactness of the sets which are involved. Further, the fact that the curve is of class $C^{(1)}$ has not been used yet, merely the continuity of the curve, so its natural to ask whether this also holds for a curve which is simply continuous, or whether I need this additional hypothesis to complete the curve.  
If anyone could either help me finish with my method, or provide me with an alternative argument for the existence of the desired partition, I would be extremely grateful. 
Also, would anything changed if we restricted the members of $\Phi$ to be open disks?
Note: I realize the title is rather vague, but I couldn't come up with something more specific (and short), so I'm open for suggestions. 
 A: $\newcommand{\Cpx}{\mathbf{C}}$Let $\alpha:[a,b] \to \Cpx$ be a continuous curve (such as a curve of class $C^{1}$), and let $\{V_{c}\}_{c \in C}$ be an arbitrary (finite or not) covering of the image $\alpha([a,b])$ by open subsets of $\Cpx$.
By continuity of $\alpha$, each preimage $U_{c} = \alpha^{-1}(V_{c}) \subset [a, b]$ is (relatively) open, and the family $\{U_{c}\}_{c \in C}$ is a covering of $[a, b]$. Replace each $U_{c}$ by its set of connected components, each of which is an open subinterval of $[a, b]$. The collection of all such intervals is also an open covering of $[a, b]$. By compactness of $[a, b]$, there exists a finite subcovering, say $\{I_{0}, I_{1}, \dots, I_{n}\}$, and the image $\alpha(I_{i})$ of each interval is contained in one of the $V_{c}$.
Now we'll build a "daisy-chain" from the $(I_{i})_{i=0}^{n}$, starting at $a$ and ending at $b$: Let $J_{0}$ be an interval containing $a$ and having maximal supremum (rightmost right-hand endpoint). Let $J_{1}$ be an interval containing $\sup I_{0}$ and having maximal supremum. Continue in this fashion until you get an interval $J_{m}$ containing $b$. (In other words, the zeroth interval $J_{0}$ contains $a$ and overlaps $J_{1}$, which overlaps $J_{2}$, etc., up to $J_{m}$ which contains $b$.) A suitable partition is
$$
t_{0} = a
  < t_{1} = \tfrac{1}{2}(\sup J_{0} + \inf J_{1})
  < \cdots
  < t_{m} = \tfrac{1}{2}(\sup J_{m-1} + \inf J_{m})
  < b = t_{m+1}.
$$
(In words, each $t_{j}$ is the midpoint of $J_{j-1} \cap J_{j}$.) Clearly $[t_{j}, t_{j+1}] \subset J_{j}$ for each $j = 0, \dots, m$, so
$$
\alpha([t_{j}, t_{j+1}]) \subset \alpha(J_{j})
$$
is contained in some $V_{c}$.
A: Suppose $K$ is a compact subset of a metric space $X$ and $\{V_\alpha\}$ is an open cover of $K$ in $X.$ Then there exists $\epsilon>0$ such that if $E\subset K$ and $\text {diam } E < \epsilon,$ then $E$ is contained in some $V_\alpha.$ Proof: Suppose not. Then for each $n \in \mathbb {N},$ there exists $E_n \subset K$ with $\text {diam } E_n < 1/n$ such that $E_n$ is contained in no $V_\alpha.$ Choose $x_n \in E_n.$ Then some subsequence $x_{n_k}$ converges to some $x_0 \in K.$ Now $x_0$ belongs to some $V_{\alpha_0}.$ Hence $B(x_0,r) \subset V_{\alpha_0}$ for some $r>0.$ Now you can verify easily that for large $k,$ $E_{n_k}\subset B(x_0,r) \subset V_{\alpha_0},$ contradiction.
Now let $\alpha :[a,b]\to \mathbb {C}$ be any continuous map (we do not need the $C^1$ condition). Then $\alpha([a,b])$ is compact. Suppose $\{V_\alpha\}$ is an open cover of $\alpha([a,b]).$ Choose $\epsilon > 0$ as above. Because $\alpha$ is uniformly continuous, there is a partion $a=t_0 < \cdots < x_n=b$ such that $\text {diam } \alpha ([t_{k-1},t_k]) < \epsilon$ for all $k.$ Then each $\alpha ([t_{k-1},t_k])$ is contained in some $V_\alpha$ as desired.
A: After an afternoon of obsessing over the problem and gesturing like a lunatic on the bus, I think I might have got it. What follows completes (I think) the argument in my question. It should be remarked that we need only assume the $\alpha$ is continuous.  
So far, we've inductively constructed sequences $\{t_k\}$ and $\{ V_k\}\subset \Phi$ such that: 
$$t_k=\min\{t\in [t_{k-1},b] : \alpha(t) \not \in V_k\}$$
and $V_{k+1}$ is some member of the cover such that $\alpha(t_k) \in V_{k+1}$. For definiteness, index the members of $\Phi$ in some way, and in each step choose the open set with the least index such that $\alpha(t_k)$ belongs to it. Obviously, these indexes need not coincide with those of the sequence, and if the sequence $\{t_k\}$ is infinite, then the sets $V_k$ will eventually start to repeat, since $\Phi$ is finite. 
Now, suppose for the sake of contradiction that this process never stops. More precisely:
$$\left ( \forall k \in \mathbb{N} \right ) \left ( t_k<b \ \mbox{and} \ \{ t\in [t_k,b] : \alpha(t) \not \in V_k \}\neq \emptyset \right )$$ 
The construction can therefore be continued indefinitely to obtain a strictly increasing sequence $\{t_k\}_{k=0}^\infty$, with $t_0=a$. 
Since this sequence is increasing and bounded above (by $b$), there exists:
$$t^*=\lim_{k\to \infty}t_k$$
Furthermore, since $\{t_k\}_{k=0}^\infty$ is strictly increasing:
$$t_k<t^*\le b \quad \forall k\in \mathbb{N}$$
Now, since $t^*\in (a,b]$ and $\Phi$ is a cover of $\gamma$, $\alpha(t^*)\in V^*$ for some $V^*$ in $\Phi$. Since $V^*$ is open, there is some $\varepsilon >0$ such that $D(\alpha (z^*),\varepsilon)\subset V^*$. By continuity, there exists $\delta>0$ such that $|t-t^*|<\delta \implies \alpha(t)\in D(\alpha(z^*),\varepsilon)$. Since $t_k \to t^*$, we can find $K$ such that $\left |t_K-t^* \right  |< \delta$. But then $\alpha(t_K),\alpha(t_{K+1})\in D(\alpha(z^*),\varepsilon)\subset V^*$, which is a contradiction since the images by $\alpha$ of consecutive members of the sequence $\{t_k\}_{k=0}^\infty$ must be in different members of the cover. 
So we conclude that, at some point, either $t_n=b$ or $\{ t\in [t_n,b] : \alpha(t) \not \in V_n \}= \emptyset$. 
In the first case, if $t_n = b$, this means that the curve "leaves" the set $V_n$ exactly at $b$. Now, $\alpha(t_n)$ must belong to some member of the cover, say $V_{n+1}$. Since $V_{n+1}$ is open, there is some $r>0$ such that $D(\alpha(b),r)\subset V_{n+1}$. By continuity, we can choose $t'\in (t_{n-1},b)$ such that $\alpha(t')\in  D(\alpha(b),r)\subset V_{n+1}$. It is clear that the partition $a=t_0<t_1<\dots<t_{n-1}<t'<t_n=b$ verifies what we want (well, not exactly, but more on that later). 
In the second case, if for some $n$, $\{ t\in [t_n,b] : \alpha(t) \not \in V_n \}=\emptyset$, then $\alpha ([t_n,b])\subset V_n$, and hence the partition $a=t_0<t_1<\dots<t_n=b$ is as desired. 
Now, what we have proved is not exactly what we wanted, since the partition verifies $\alpha([t_{k-1},t_k)) \subset V_k$ and $\alpha([t_n,b])\subset V_n$ for all $k=1,\dots,n-1$, instead of $\alpha([t_{k-1},t_k])$ for $k=1,\dots,n$. For my original purposes, this is sufficient, but in lieu of completeness I'll indicate how to remedie this: 
First, we observe that $\alpha(t_k) \in \partial V_k$. If this were not so, since $\alpha(t_k) \not \in V_k$ and $\overline{V_k}=V_k \cup \partial V_k$, then $t_k \not \in \overline{V_k}$. Since $\overline{V_k}$ is closed and $\{t_k\}$ is compact, $d(t_k,\overline{V_k})=\eta_k>0$. But then, by continuity, we can find $t'\in (t_{k-1},t_k)$ such that $\alpha(t')\in D(\alpha(t_k),\eta_k)$, and hence $\alpha(t')\not \in V_k$ by our choice of $\eta_k$, which contradicts the definition of $t_k$ since $t_{k1}<t'<t_k$. It follows therefore that $\alpha(t_k) \in \overline{V_k}$ for all $k$, $k=1,\dots,n$. 
Now, for each $k$ with $1\le k\le n$, $\alpha(t_k)\in V_{k+1}$, which is open, so there exist positive numbers $\eta_k$ such that $D(\alpha(t_k),\eta_k)\subset V_{k+1}$. Again, by continuity, we can find $t'_k\in(t_{k-1},t_k)$ such that $\alpha([t'_k,t_k))\subset D(\alpha(t_k),\eta_k)\subset V_{k+1}$. Since $t_k<t'_{k+1}<t_{k+1}$, it follows that $\alpha([t_k,t'_{k+1}])\subset V_{k+1}$. Hence $\alpha([t'_k,t'_{k+1}])\subset V_{k+1}$. 
Finally, since $\alpha([a,t'_1])\subset V_1$ and $\alpha([t'_n,b])\subset V_n$, it is clear that, after reindexing, the partition $a<t'_1<t'_2<\dots<t'_n<b$ has the desired properties. 
