First we use that
$$
\int_X g\,\mathrm d\mu=\int_0^\infty \mu\left(\{g\geq t\}\right)\,\lambda(\mathrm dt)
$$
for any non-negative measurable function $g$. Using this with $g=f^2$ we get
$$
\int_X f^2\,\mathrm d\mu=\int_0^\infty \mu\left(\{f\geq \sqrt{t}\}\right)\,\lambda(\mathrm dt)=2\int_0^\infty x\cdot \mu\left(\{f\geq x\}\right)\,\lambda(\mathrm dx)
$$
by change of variables with $x=\sqrt{t}$.
Now $x\mapsto x\cdot \mu\left(\{f\geq x\}\right)$ is measurable and $x\mapsto \mu\left(\{f\geq x\}\right)$ is decreasing, and so
$$
\sum_{n=1}^\infty n\cdot \mu\left(\{f\geq n-1\}\right)1_{(n-1,n]}(x)\geq x\cdot \mu\left(\{f\geq x\}\right)\geq \sum_{n=1}^\infty (n-1)\cdot \mu\left(\{f\geq n\}\right)1_{(n-1,n]}(x)
$$
and integrating this with respect to $\lambda$ yields
$$
\int_0^\infty x\cdot \mu\left(\{f\geq x\}\right)\,\lambda(\mathrm d x)\leq \sum_{n=1}^\infty \left(\int_0^\infty n\cdot\mu\left(\{f\geq n-1\}\right)1_{(n-1,n]}(x)\,\lambda(\mathrm d x)\right)\\
=\sum_{n=1}^\infty n\cdot\mu\left(\{f\geq n-1\}\right)=\sum_{n=0}^\infty (n+1)\cdot\mu\left(\{f\geq n\}\right).
$$
Now this last sum is finite if and only if $\sum_{n=1}^\infty n\cdot\mu\left(\{f\geq n\}\right)<\infty$ (by the limit comparison test).