As far as I can tell the following argument given in the case of $\mathbb{P}^n$ from Okonek-Schneider-Spindler "Vector bundles on complex projective spaces" works as well. If someone could check this over I'd be grateful, as I'd be surprised if Uhlenbeck-Yau missed it.

First show that if $f:E_1\to E_2$ is a homomorphism of stable bundles of the same slope then $f$ has the "expected rank," i.e. it is either injective or generically surjective. For if $I$ is the image sheaf and its rank is smaller than expected, we have by stability

$$\mu(I) > \mu(E_1) = \mu(E_2) > \mu(I),$$

a contradiction.

Next suppose $f:E_1\to E_2$ is a nonzero homomorphism of stable bundles of the same rank and $c_1$. By the previous paragraph, $f$ is injective. Then the induced map $\det E_1\to \det E_2$ is also injective; this is a map of line bundles of the same $c_1$, so it is in fact an isomorphism. But then $f$ originally must have been an isomorphism.

In particular, any nonzero endomorphism $f:E\to E$ of a stable bundle is an isomorphism. Now choose a point $x\in X$, and look at the action $f_x : E_x\to E_x$ on the fiber. Choose an eigenvalue $\lambda\in \mathbb{C}$ of $f_x$. Then $f-\lambda I$ is either zero or an isomorphism, but it isn't an isomorphism at $x$ so we conclude $f = \lambda I$.

As far as I can tell the following argument given in the case of $\mathbb{P}^n$ from Okonek-Schneider-Spindler "Vector bundles on complex projective spaces" works here as well. If someone could check this over I'd be grateful, as I'd be surprised if Uhlenbeck-Yau missed it.

First show that if $f:E_1\to E_2$ is a homomorphism of stable bundles of the same slope then $f$ has the "expected rank," i.e. it is either injective or generically surjective. For if $I$ is the image sheaf and its rank is smaller than expected, we have by stability

$$\mu(I) > \mu(E_1) = \mu(E_2) > \mu(I),$$

a contradiction.

Next suppose $f:E_1\to E_2$ is a nonzero homomorphism of stable bundles of the same rank and $c_1$. By the previous paragraph, $f$ is injective. Then the induced map $\det E_1\to \det E_2$ is also injective; this is a map of line bundles of the same $c_1$, so it is in fact an isomorphism. But then $f$ originally must have been an isomorphism.

In particular, any nonzero endomorphism $f:E\to E$ of a stable bundle is an isomorphism. Now choose a point $x\in X$, and look at the action $f_x : E_x\to E_x$ on the fiber. Choose an eigenvalue $\lambda\in \mathbb{C}$ of $f_x$. Then $f-\lambda I$ is either zero or an isomorphism, but it isn't an isomorphism at $x$ so we conclude $f = \lambda I$.