= Solution
Choose $P\in X(k)$ and put $D=(g+1)P$. The <Riemann-Roch theorem> gives
$$
\ell(D)=\deg D+1-g+\ell(K-D)=2+\ell(K-D)\ge2.
$$
As constants supply only a one-dimensional subspace, the <Riemann-Roch space> $L(D)$ contains a nonconstant <rational function> $f$. It has no poles except possibly at $P$. It must have a pole, since a global <regular function> on an irreducible <projective variety> is constant. Let its pole order at $P$ be $m$; then $1\le m\le g+1$.
On the open set where $f$ is regular, define $\phi=[f:1]$. At its pole use the other affine chart, $\phi=[1:1/f]$. The <local ring of a smooth algebraic curve> is a <discrete valuation ring>, and the positive valuation of $1/f$ at $P$ makes the latter expression regular there. The two expressions agree on their overlap, so they give a regular map $X\to\mathbb P^1$.
This map is nonconstant. It is proper because its source is projective and its target is separated, so its image is closed; a nonconstant image in the projective line must be the whole line. Each finite-value fibre consists of zeros of a nonzero <rational function> $f-a$, and the infinity fibre is the finite pole set. Hence every fibre is finite, giving a <quasi-finite morphism>. The proper quasi-finite criterion therefore makes $\phi$ a <finite morphism>.
Finally its pullback of the infinity divisor is exactly the pole divisor $mP$. The <degree of a morphism of curves> equals the degree of this pullback, or equivalently the function-field extension degree $[k(X):k(f)]$. Thus
$$
\boxed{\deg\phi=m\le g+1.}
$$
The argument includes inseparable maps: the degree here is the full field-extension degree, not just the separable degree. This proves the <gonality bound from Riemann-Roch> with an explicitly constructed finite regular map.
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