Well, the discrete version of this question (where $Tf(m) = \sum K(m,n)f(n)$ maps $l^\infty$ to $l^\infty$) isn't so hard. That should give you an intuition for why it's true. I don't see any slick way to handle the continuous case; I guess you could do it first when $K$ is the characteristic function of a rectangle, then when $K$ is a finite linear combination of such things, then when $K$ is the characteristic function of any measurable set, then when $K$ is any simple function, and then finally for general $K$.
Incidentally, the reverse inequality is easier. For $h \in L^1$ let $H$ be the bounded linear functional on $L^\infty$ given by integrating against $h$. Then for any $\epsilon > 0$ we can find $h$ with $\|h\|_1 = 1$ such that $\|H \circ T\| \geq \|T\| - \epsilon$. But $$(H \circ T)(f) = \int h(x)\int K(x,y) f(y)dydx = \int\left(\int h(x)K(x,y)dx\right)f(y)dy$$ is integration against the function $\int h(x)K(x,y)dx$, and the $L^1$ norm of this function is at most $\int\int |h(x)||K(x,y)|dxdy = \int |h(x)|\big(\int |K(x,y)|dy\big)dx \leq \|\int|K(\cdot,y)|dy\|_\infty$. At least that works if $K$ is integrable so you can change order of integration.