$\|T_af-f\|_\infty\rightarrow 0,\ a\rightarrow 0\Rightarrow\ f$ has uniformly continuous representative

Question

Let $f\in L^\infty(\mathbb{R}^d)$ (where the measure we're working with is the Lebesgue measure) and I have to prove the following equivalence :
$f$ has a uniformly continuous representative if and only if $\|T_af-f\|_\infty \rightarrow 0$ as $a\rightarrow 0$
where $T_af$ is defined almost everywhere by $T_af(x)=f(a+x)\ \forall a,x\in \mathbb{R}^d$.
I think I managed to prove the forward implication but I have no idea how to prove the inverse one. Here is my work :
Let $g$ be a uniformly continuous representative of $f$ and $\epsilon >0$. Let $\eta>0$ such that $|g(x)-g(y)|<\epsilon\ \forall x,y\in \mathbb{R}^d,\ \|x-y\|<\eta$.
Then $\forall a\in \mathbb{R}^d\ (\|a\|<\eta\Rightarrow |g(x+a)-g(x)|<\epsilon\ \forall x\in \mathbb{R}^d$).
So $\forall a\in \mathbb{R}^d$ such that $\|a\|<\eta$ we have:
$\|T_af-f\|_\infty=\sup_{x\in \mathbb{R}^d}|g(x+a)-g(x)|<\epsilon$ and so $\|T_af-f\|_{\infty}\rightarrow 0$ as $a\rightarrow 0$.
I would be grateful for any help with the inverse implication or if you could check my answer for the forward implication.