Suppose that $R$ is a torsion-free subsheaf (of positive rank) in another torsion-free sheaf $S$, on a smooth complex projective variety $X$.
If $S$ is (slope) semistable, is it true that the degree of $R$ is always less than or equal to that of $S$? If so, why?
This is not a question about how to define the degree, but if you believe the method of definition affects the answer, then please explain.
Also, I prefer not to make assumptions about the dimension of $X$.
A reference to a specific result in a book or paper would be appreciated in lieu of an explanation (but only if you are sure about the reference and the specificity).
(Note: The first sentence of Krish's answer below refers to an earlier version of the question, in which stability was not required.)
Even for vector bundles this is not true. Take $X = \mathbb{P}^1, S = \mathcal{O}(1) \oplus \mathcal{O}(-1), R = \mathcal{O}(1).$
Note that, we always have the following relation deg$(R) +$ deg$(S/R) =$ deg$(S)$, where degree is defined with respect to a fixed ample divisor on $X.$ So deg$(R) \leq$ deg$(S) \Leftrightarrow$ deg$(S/R) \geq 0.$ Even if deg$(S) \geq 0,$ in general it can happen that deg$(S/R) < 0$ (see the above example).
But if $S$ is semi-stable, then it is true that deg$(R) \leq$ deg$(S)$, provided deg$(S) \geq 0$. If deg$ (S ) < 0$, then it is not true in general. Take $ X = \mathbb {P}^n, n \geq 1$ and consider the following exact sequence $$ 0 \rightarrow \mathcal {O}(-1) \rightarrow \mathcal {O}(-1) \oplus \mathcal {O}(-1) \rightarrow \mathcal {O}(-1) \rightarrow 0. $$