This is an exercise straight out of Lang's Undergraduate Analysis (Ch 2, sec 4, 1).
Here is my attempt at a proof, please let me know if I am on the right track.
Proof: We will show that $f$ is uniformly continuous on $\mathbb{R}$. Let $\epsilon >0$ and consider $t\in \mathbb{R}$. Choose $\delta$ such that $\delta<\frac{\epsilon|t|}{|f(t)|}$. Now, for $x,y\in\mathbb{R}$, if $|x-y|<\delta$, then we have
$$|x-y|<\frac{\epsilon|t|}{|f(t)|}$$ $$|f(t)||x-y|<\epsilon|t|$$ $$|xf(t)-yf(t)|<\epsilon|t|$$ $$|tf(x)-tf(y)|<\epsilon|t|$$ $$|t||f(x)-f(y)|<\epsilon|t|$$ $$|f(x)-f(y)|<\epsilon$$.
I think you're proof is correct, just pay attention to the case where the function $f(x)=0 \; \forall x \in \mathbb R$, because in that case the claim is obviously true, but your demonstration is not valid because $\delta$ can't be defined in that way (you can't divide for $0$). I would specify that $f(t)$ should be different than $0$ and i would say that if i can't find such a $t$ then i have $f(x)=0 \; \forall x \in \mathbb R$. Also note that $\delta>0$ is required, so we can't have $t=0$. Finally we can't have the unpleasant case in which $f(x)=0, \forall x \in \mathbb R/ \{0 \}$ and $f(0)\neq 0$ in fact $f(0)=f(t \bullet 0)=t\bullet f(0) \; \forall t \in \mathbb R, \rightarrow f(0)=0 $. We can then always find a $t \in \mathbb R$ such that $t\neq 0$ and $f(t)\neq 0$ or we are in the simple case of $f(x)=0 \; \forall x \in \mathbb R$.