I assume that the limit is taken as $y$ goes down to $0$ from the positive side. Let $t=\frac{\lambda-x}{y}$. The integral becomes
$$V(x,y)=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{F(x+yt)}{1+t^2}dt.$$
Since $F$ is bounded, there exists $M>0$ such that $|F(z)|\leq M$ at all $z\in \Bbb{R}$.
Fix $x\in\Bbb{R}$ and $\epsilon>0$. There exists $L(\epsilon)>0$ such that $$\int_{L(\epsilon)}^\infty\frac{1}{1+t^2}dt<\frac{\pi\epsilon}{8M}.\tag{1}$$
For example, you may take $L(\epsilon)=\frac{8M}{\pi\epsilon}$.
Recall that $F$ is continuous. Pick $\delta>0$ so small that $\big\vert F(z)-F(x)\big\vert<\frac{\epsilon}{2}$ for all $z\in\Bbb{R}$ such that $|z-x|<L(\epsilon)\delta$. Thus, for $0<y<\delta$,
$$\big\vert V(x,y)-F(x)\big\vert\leq \frac{1}{\pi}\int_{-L(\epsilon)}^{L(\epsilon)}\frac{\big|F(x+yt)-F(x)\big|}{1+t^2}dt+\frac{1}{\pi}\int_{L(\epsilon)}^\infty\,\frac{2M}{1+t^2}dt+\frac{1}{\pi}\int_{-\infty}^{-L(\epsilon)}\frac{2M}{1+t^2}dt.$$
That is,
$$\big\vert V(x,y)-F(x)\big\vert\leq \frac{1}{\pi}\int_{-L(\epsilon)}^{L(\epsilon)}\frac{\frac{\epsilon}{2}}{1+t^2}dt+\frac{4M}{\pi}\int_{L(\epsilon)}^\infty\,\frac{1}{1+t^2}dt.$$
From (1), we have
$$\big\vert V(x,y)-F(x)\big\vert< \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\frac{\epsilon}{2}}{1+t^2}dt+\frac{4M}{\pi}\left(\frac{\pi\epsilon}{8M}\right)=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$
Thus,
$$\lim_{y\searrow0}V(x,y)=F(x).$$
