That is not true. Let $k$ be an algebraically closed field, e.g., the field of complex numbers. Let $X$ be a smooth, projective $k$-curve of genus $g\geq 2.$ Choose a $k$-basis of the following $g$-dimensional $k$-vector space, $$e_1,\dots,e_g\in \text{Ext}^1_k(\mathcal{O}_X,\mathcal{O}_X) = H^1(X,\mathcal{O}_X).$$ Each element $e_i$ gives a Yoneda extension class of a short exact sequence, $$e_i: \ \ 0 \to \mathcal{O}_X \xrightarrow{q_i} E_i \xrightarrow{p_i} \mathcal{O}_X \to 0.$$ Each element $E_i$ is semistable (not polystable) of slope $0$. Thus the direct sum $F= E_1\oplus \dots \oplus E_g$ is also semistable of slope $0$. There is a short exact sequence, $$0 \to \bigoplus_{i=1}^g \mathcal{O}_X \xrightarrow{\oplus q_i} \bigoplus_{i=1}^g E_i \xrightarrow{\oplus p_i} \bigoplus_{i=1}^g \mathcal{O}_X \to 0. $$ For the subbundle, there is also a "summing" surjective $\mathcal{O}_X$-module homomorphism, $$\Sigma:\bigoplus_{i=1}^g \mathcal{O}_X \to \mathcal{O}_X.$$ Denote the kernel by $K=\text{Ker}(\Sigma).$ Define $E$ to be the quotient of $F$ by $K.$ Thus, there is a short exact sequence, $$e:\ \ 0 \to \mathcal{O}_X \xrightarrow{q} E \xrightarrow{\oplus p_i} \bigoplus_{i=1}^g \mathcal{O}_X \to 0. $$ Again, since the first and third terms are semistable of slope $0$, also the middle term $E$ is semistable of slope $0.$
Now, for every nonzero element $a=(a_1,\dots,a_g)\in k^g,$ consider the subbundle $E_a$ of $E$ that is the pullback by the following map, $$a:\bigoplus_{i=1}^g \mathcal{O}_X \xrightarrow{(a_1,\dots,a_g)} \mathcal{O}_X.$$ In other words, there is a commutative diagram of short exact sequences, $$\begin{array}{cccccccccc} e_a: & 0 & \to & \mathcal{O}_X & \xrightarrow{q} & E_a & \to & \mathcal{O}_X & \to & 0 \\ & & & \text{Id}~\downarrow & & \downarrow{f_a} & & \downarrow{a} \\ e: & 0 & \to & \mathcal{O}_X & \xrightarrow{q} & E & \xrightarrow{\oplus p_i} & \bigoplus_{i=1}^g \mathcal{O}_X & \to & 0 \end{array}.$$
Finally, for every pair of elements $a,b\in k^g\setminus \{0\}$ that are not related by scaling, the semistable sheaves $E_a$ and $E_b$ are not isomorphic. Associated to the short exact sequence $e_a$, consider the long exact sequence of cohomology, $$0 \to H^0(X,\mathcal{O}_X) \to H^0(X,E_a) \to H^0(X,\mathcal{O}_X) \xrightarrow{\partial_a} H^1(X,\mathcal{O}_X).$$ Since $a$ is nonzero, the first map above is an isomorphism, and the last map above is an injection. Thus, the subsheaf $q(\mathcal{O}_X)$ is the intrinsic subsheaf $H^0(X,E_a)\otimes_k \mathcal{O}_X,$ the cokernel by this intrinsic sheaf is isomorphic to $\mathcal{O}_X,$ and the connecting map $\delta$ is identified with the intrinsic connecting map, $$\partial_{E_a}:H^0(X,E_a/H^0(X,E_a)\otimes_k\mathcal{O}_X) \to H^1(X,H^0(X,E_a)\otimes_k \mathcal{O}_X), \text{ i.e.,}$$ $$\partial_{E_a}:H^0(X,E_a/H^0(X,E_a)\otimes_k\mathcal{O}_X) \to H^0(X,E_a)\otimes_k H^1(X,\mathcal{O}_X).$$ Of course the image of $\partial_a$ in $H^1(X,\mathcal{O}_X)\setminus \{0\}$ gives a well-defined and intrinsic element in the projective space, $$\mathbb{P} H^1(X,\mathcal{O}_X) \cong \mathbb{P}^{g-1}_k.$$ Thus, the sheaf $E_a$ is isomorphic to $E_b$ if and only if the elements $[a],[b]\in \mathbb{P} H^1(X,\mathcal{O}_X)$ are equal. For $g\geq 2,$ there are infinitely many $k$-points of $\mathbb{P}^{g-1}_k.$ Thus, there are infinitely many isomorphism classes of semistable subsheaves of $E$ of slope $0$.