Here is a complete proof: as remarked in the answer by Norondion, we can reduce to
the case when $C\_1 \rightarrow C\_2$ is generically separable, i.e. $k(C\_1)$ is separable
over $k(C\_2)$. Let $A \subset k(C\_1)$ be a finite type $k$-algebra consisting of the regular
functions on some non-empty affine open subset $U$ of $C\_2$ (it doesn't matter which one
you choose), so that $k(C\_2)$ is the fraction field of $A$.

By the primitive element theorem, we may write $k(C\_1) = k(C\_2)[\alpha]$, where
$\alpha$ satisfies some polynomial $f(\alpha) = \alpha^n + a_{n-1}\alpha^{n-1} + \cdots
+ a_1 \alpha + a_0 = 0,$ for some $a_i$ in $K(C\_2)$.

Now the $a_i$ can be written as fractions involving elements of $A$, i.e. each
$a_i = b_i/c_i$ for some $b_i,c_i \in A$ (with $c_i$ non-zero). We may replace
$A$ by $A[c\_0^{-1},\ldots,c\_{n-1}^{-1}]$ (this corresponds to puncturing $U$ at
the zeroes of the $c_i$), and thus assume that in fact the $a_i$ lie in $A$.

The ring $A[\alpha]$ is now integral over $A$, and of course has fraction field equal
to $k(C_2)[\alpha] = k(C_1)$. It need not be that $A[\alpha]$ is integrally closed,
though. We are going to shrink $U$ further so we can be sure of this.

By separability of $k(C_1)$ over $k(C_2)$, we know that the discriminant $\Delta$
of $f$ is non-zero, and so replacing $A$ by $A[\Delta^{-1}]$ (i.e. shrinking $U$
some more) we may assume that $\Delta$ is invertible in $A$ as well.

It's now not hard to prove that $A[\alpha]$ is integrally closed over $A$. Thus
$\text{Spec }A[\alpha]$ is the preimage of $U$ in $C_1$ (in a map of smooth curves,
taking preimages of an affine open precisely corresponds to taking the integral closure
of the corresponding ring).

In other words, restricted to $U \subset C_2$, the map has the form
$\text{Spec }A[\alpha] \rightarrow \text{Spec }A,$ or, what is the same,
$\text{Spec }A[x]/(f(x)) \rightarrow \text{Spec A}$.

Now if you fix a closed point $\mathfrak m \in \text{Spec }A,$ the fibre over this point
is equal to $\text{Spec }(A/\mathfrak m)[x]/(\overline{f}(x)) = k[x]/(\overline{f}(x)),$
where here $\overline{f}$ denotes the reduction of $f$ mod $\mathfrak m$.
(Here is where we use that $k$ is algebraically closed, to deduce that $A/\mathfrak m = k,$
and not some finite extension of $k$.)

Now we arranged for $\Delta$ to be in $A^{\times}$, and so $\bar{\Delta}$ (the reduction
of $\Delta$ mod $\mathfrak m$, or equivalently, the discriminant of $\bar{f}$)
is non-zero, and so $k[x]/(\bar{f}(x))$ is just a product of copies of $k$,
as many as equal to the degree of $f$, which equals the degree of $k(C_1)$ over
$k(C_2)$. Thus $\text{Spec }k[x]/(\bar{f}(x))$ is a union of that many points,
which is what we wanted to show.