Using $d_\infty(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$, prove the above.
I think I have a proof but it seems weird and I think it might be wrong.
First, note that all functions in $C([0,1])$ are bounded since they are continuous functions on a compact set. Also, for any $f,g\in C([0,1])$ , $h=f-g\in C([0,1])$ since it is the difference of continuous functions. Fix $\epsilon>0$ and pick $f,g$ so that $h=f-g$ is bounded by some $\delta<\epsilon$ [this is the part that seems too simple I think] , that is, $|f(x)-g(x)|<\delta$ for all $x$ . Taking the supremum of the left hand side we can see that $d_{\infty}(f,g)<\delta$ . Now $$\left\Vert I(f)-I(g)\right\Vert = \sup_{x\in[0,1]}\left|\int_{0}^{x}f(t)dt-\int_{0}^{x}g(t)dt\right| = \sup_{x\in[0,1]}\left|\int_{0}^{x}\left[f(t)-g(t)\right]dt\right| = \sup_{x\in[0,1]}\left|\int_{0}^{x}h(t)dt\right| $$ and since $h(t)$ is bounded by $\delta$ on $[0,1]$ we have for all $x\in[0,1]$ $$\int_{0}^{x}h(t)dt \leq \delta(1-0) = \delta < \epsilon$$ taking the supremum of the left hand side gives the desired result.
It seems too simple, like maybe I'm missing something. Am I?
If you ask about uniform continuity of $I$ as your title suggests, here is a proof.
I will denote $\|f\|_\infty=d_\infty(f,0)=\sup |f(x)|$. This is a standard notation for the uniform norm. Your question presents $C([0,1])$ as a metric space. This is true, but there is much more structure: it is a normed vector space.
Note that $I$ is a linear operator. Now for every $f$ $$ |I(f)(x)|=|\int_0^xf(t)dt|\leq\int_0^x|f(t)|\leq \|f\|_\infty\int_0^xdt\leq \|f\|_\infty $$ for all $x$ So $$ \|I(f)\|_\infty\leq \|f\|_\infty $$ This means that $I$ is bounded and implies in particular $$ d_\infty(I(f),I(g))=\|I(f)-I(g)\|_\infty\leq \|f-g\|_\infty=d_\infty(f,g). $$ Now you see that $I$ is $1$-Lipschitz. In particular, it is uniformly continuous.
Note: every bounded linear operator on a normed vector space is uniformly continuous.