Hartshorne Exercise IV.1.2.
Again let $X$ be a curve, and let $P_1,…,P_r\in X$ be points. Then there is a rational function $f\in \Bbb K(X)$ having poles (of some order) at each of the $P_i$, and regular elsewhere.
For large $n$ consider the divisor $D=n\sum_iP_i$. We have $h^0(X,\OO_X(D))=nr+1-g$ for $n>>0$. For each $1\le i\le r$, consider the divisor $D_i=D-nP_i$. Then $H^0(X,\OO_X(D_i))\subset H^0(X,\OO_X(D))$ is a proper subspace that corresponds to the subspace of $\Bbb K(X)$ consisting of rational functions that are regular away from $P_1,…\hat{P_i},…P_n$ with poles of order at most $n$ at each of those points. Then $H^0(X,\OO_X(D))-\bigcup_{i=1}^rH^0(X,\OO_X(D_i))\ne \varnothing$, and any section in there corresponds to a rational function that is regular away from $P_1,…,P_n$ with poles of order at most $n$ at each $P_i$. I claim that such a function has a pole at each of the points. If it were missing a pole at some $P_i$ then it would be regular away from $D_i$ and hence in $H^0(X,\OO_X(D_i))$ which is a contradiction.
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