= Solution
Take the <compact Riemann surface> $X=\mathbb P^1$ and the <holomorphic vector bundle> $E=\mathcal O_X\oplus\mathcal O_X$. For every positive integer $m$, use homogeneous coordinates $[Z_0:Z_1]$ and the map
$$
\mathcal O_X(-m)\longrightarrow\mathcal O_X^{\oplus2},\qquad
s\longmapsto(Z_0^m s,Z_1^m s).
$$
The two sections $Z_0^m,Z_1^m$ never vanish together on $\mathbb P^1$. Thus this map has rank one at every point; locally one of its two components is invertible in suitable line-bundle frames, so its image is a genuine <holomorphic subbundle> and its quotient is a <locally free sheaf>. More explicitly, there is an <exact sequence>
$$
0\longrightarrow\mathcal O_X(-m)
\xrightarrow{(Z_0^m,Z_1^m)}\mathcal O_X^{\oplus2}
\xrightarrow{(-Z_1^m,Z_0^m)}\mathcal O_X(m)
\longrightarrow0.
$$
Exactness follows fibrewise from the absence of a common zero and hence also locally as a sequence of vector bundles. The image $F_m$ is isomorphic to $\mathcal O_X(-m)$ and has rank one, so
$$
\boxed{\mu(F_m)=\deg\mathcal O_X(-m)=-m\longrightarrow-\infty.}
$$
Thus \b[there need not be a lower bound for subbundle slopes], even for a fixed trivial rank-two bundle. Merely multiplying a section of one fixed subline by a function with zeros would not suffice: that could give only a subsheaf with a torsion quotient. The nowhere-simultaneously-zero pair above avoids that problem and realizes <unbounded negative degrees of line subbundles>.
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