Let $D$ be a strictly positive divisor defined on a compact Riemann Surface such that $\operatorname{dim} \mathfrak{L}(D)=1+\operatorname{deg} D$. There exists a point $p \in X$ such that $\operatorname{dim} \mathfrak{L}(p)=2$?
Firstly, we observe that $\operatorname{codim} \mathfrak{L}(p) \le \operatorname{deg}D-1$ (this is the consequence of the exercise V.3.I by Miranda), so, using the hypothesis and the definitions, we obtain $\operatorname{dim} \mathfrak{L}(p) \ge 2$. I complete the proof if I can show that $\operatorname{codim} \mathfrak{L}(p) = \operatorname{deg}D-1$, that is if I can show the existence of a divisor $D$' such that $\operatorname{deg}D$'$=\operatorname{deg}D-1$ and $D-D$'$=p$.
Now, $D$ is linearly equivalent to a divisor $p_1+\dots+p_{\operatorname{deg}(D)}$. May I take the divisor $D$'$=p_2+\dots+p_{\operatorname{deg}(D)}$ to finish the proof?
I really did not understand your argumentation and therefore I cannot comment on that. For instance, you probably meant something else here:"that is if I can show the existence of a divisor $D$' such that $\operatorname{deg}D$'$=\operatorname{deg}D-1$ and $D-D$'$=p$." Because there are obviously divisors of that form, just take $D'$ to be $D-p$, and clearly that does not imply anything.
Anyway, I think the crucial observation here is that for any divisor $D$, the difference $\operatorname{dim}\mathcal{L}(D)- \operatorname{dim}\mathcal{L}(D-p)=0\mbox{ or } 1$. (That was either an exercise or a lemma in Miranda's book). Using this fact you can easily solve the rest.