= Solution
Let $C$ be the <smooth algebraic curve>. A nonzero regular section of a <line bundle> has an effective zero divisor, so its <degree of a line bundle> is nonnegative. Degrees add under tensor products and change sign on taking the dual. We use genuine <line subbundles>, so their quotients in $E$ are <locally free>. Both bounds can be proved constructively.
For the upper bound, take a rational frame of $E^*$ over the <function field> of $C$. Its two members are <rational sections> with only finitely many poles. Choose an <effective divisor> $D$ large enough to bound the pole orders of both <rational sections>. They become <global sections> of $E^*\otimes\mathcal O_C(D)$ and define a <sheaf> map
$$
E\longrightarrow\mathcal O_C(D)\oplus\mathcal O_C(D)
$$
which is an isomorphism at the <generic point>. It is therefore <injective> as a <sheaf> map: a section in its <kernel> would vanish generically in a <locally free sheaf>, which is a <torsion-free sheaf>. For any <line subbundle> $L\subset E$, at least one of its two component maps into $\mathcal O_C(D)$ is nonzero. Such a map is a nonzero <global section> of $\mathcal O_C(D)\otimes L^{-1}$. Its zero <divisor on an algebraic curve> is effective, so
$$
\boxed{\deg L\le\deg D.}
$$
This <upper degree bound for line subbundles on a smooth curve> depends only on a choice made for $E$, and applies to every one of its <line subbundles>.
For the lower-bound question we will actually construct subbundles $\mathcal O_C(-mp_0)$ for every sufficiently large integer $m$, where $p_0$ is a fixed point. First prove the <globally generated vector bundle> property rather than assuming that negative-degree inclusions automatically are subbundles.
Apply the upper-bound argument to the rank-two <vector bundle> $F=E^*\otimes K_C$, where $K_C$ is the <canonical bundle>. Let $B$ be its upper degree bound. Choose $m$ with $m-1>B$. For any point $p\in C$, a nonzero <global section> of $F(-mp_0+p)$ would give a nonzero map
$$
\mathcal O_C(mp_0-p)\longrightarrow F.
$$
Its image can be saturated to a <saturated line subbundle on a smooth curve> $M\subset F$. Locally the <local ring of a smooth algebraic curve> is a <discrete valuation ring>: dividing out the common power of a uniformizer makes the image a primitive rank-one summand, and its quotient is free. Thus the saturation is indeed a subbundle. The original map gives a nonzero <global section> of $M\otimes\mathcal O_C(-mp_0+p)$ and consequently $\deg M\ge m-1$, contradicting the upper bound. Hence $H^0(F(-mp_0+p))=0$ for every $p$. By <Serre duality>,
$$
H^1(E(mp_0-p))\cong H^0(E^*\otimes K_C(-mp_0+p))^*=0.
$$
The <exact sequence>
$$
0\longrightarrow E(mp_0-p)\longrightarrow E(mp_0)
\longrightarrow E(mp_0)|_p\longrightarrow0
$$
and its <long exact sequence in sheaf cohomology> show that evaluation $H^0(E(mp_0))\to E(mp_0)|_p$ is onto for every $p$. This proves the <global generation of positive twists on a smooth curve> with one bound on $m$ that works at every point.
Write $W=H^0(E(mp_0))$ and $N=\dim W$. The evaluation map $W\otimes\mathcal O_C\to E(mp_0)$ is <surjective> with <kernel> bundle of rank $N-2$. Consider the incidence subset
$$
Z=\{(p,s)\in C\times W:s(p)=0\}.
$$
It is the total space of this <kernel> bundle, so $\dim Z=1+(N-2)=N-1$. Its projection into $W$ is closed because $C$ is proper, and its image has dimension at most $N-1<N$. Choose $s$ outside this proper closed subset. It vanishes nowhere. In a local trivialization at any point, one component of $s$ is a unit, so it can be completed to a basis. Therefore $\mathcal O_C\xrightarrow{s}E(mp_0)$ is a subbundle with a <locally free> rank-one quotient. Twisting back gives
$$
\boxed{\mathcal O_C(-mp_0)\subset E,\qquad\deg\mathcal O_C(-mp_0)=-m\longrightarrow-\infty.}
$$
This proves the <unbounded negative degrees of line subbundles>. The nowhere-vanishing construction guarantees a subbundle at every point, including the zeros that might otherwise have occurred in a generic meromorphic inclusion. If the curve has several connected components, apply these constructions on each component and add the degrees; the same upper-bound and no-lower-bound conclusions follow.
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