First, assume that $K$ is continuous with compact support. Then using
$$|(f\star K)(x)-(f\star K)(x')|\leqslant\int_\mathbb R|f(y)|\cdot |K(x-y)-K(x'-y)|\mathrm dy,$$
boundedness of $f$, and uniform continuity of $K$, we get what we want.
Then we conclude by a density argument: if $K_n\to K$ in $L^1$, then $(f\star K_n)_n$ converges uniformly on the real line to $f\star K$. Indeed, we have
$$|f\star K_n(x)-f\star K(x)|\leqslant \left|\int_\mathbb Rf(x-t)K_n(t)\mathrm dt-\int_\mathbb Rf(x-t)K(t)\mathrm dt\right|\leqslant\lVert f\rVert_\infty\lVert K_n-K\rVert_1.$$
Here is another proof for the continuity of $\chi_A*\chi_B$. Let us in fact show, more generally, that if $f$ is integrable on $\mathbb R$ and $g$ is measurable and bounded, then $f*g$ is a continuous function. (Then take $f:=\chi_A$ and $g:=\chi_B$, of course).
The function $f*g(x)=\int_{\mathbb R} f(x-y)g(y)\, dy$ is clearly well-defined at every point $x\in\mathbb R$. Moreover, one may write
$$f*g(x)=\int_{\mathbb R} \tau_x f(y) g(y)\, dy\, ,$$
where $\tau_xf(y)=f(x-y)$. In an even more formal way,
$$f*g(x)=\Phi_g (\tau_x f)\, ,$$
where $\Phi_g :L^1(\mathbb R)\to \mathbb R$ is the linear functional defined by $\Phi_g(u)=\int_{\mathbb R} u(y)g(y)\, dy$. This makes sense, and $\Phi_g$ is a continuous linear functional on $L^1(\mathbb R)$, because $g$ is bounded.
So, to prove that $f*g$ is a continuous function, you only have to check that the map $x\mapsto \tau_x f$ is continuous from $\mathbb R$ into $L^1(\mathbb R)$.
But now, there is no miracle: you need to use some approximation argument, just like in the other answers already given to your question. For example, you can first check that the map $x\mapsto \tau_x f$ is indeed continuous if $f$ is continuous and compactly supported (this should cause no difficulty); and then, taking a sequence $(f_n)$ of continuous and compactly supported functions tending to $f$ with respect to the $L^1$ norm, observe that $\tau_xf_n\to \tau_x f$ uniformly on $\mathbb R$ as $n\to\infty$, because $\Vert \tau_x f_n-\tau_x f\Vert_{L^1}=\Vert f_n-f\Vert_{L^1}$.
(A proof can be found on Real and Complex Analysis, 3rd Edition, theorem 9.5. In fact, this function is uniformly continuous from $\mathbb{R}$ into $L^p(\mathbb{R})$ where $1 \le p < \infty$)
Best Answer
Exactly you can find your answer here: Part "d", Proposition "2.39", Page "52", Chapter 2, A course in Abstract Harmonic Analysis, Gerald b. Folland.
My question: When $d=1$ how you can say $C(\Bbb{R})\subset L^\infty(\Bbb{R})$? Are you sure the function $f(x)=x$ is in $L^\infty(\Bbb{R})$?