If $P\in \mathbb P^2$ is an arbitrary point, any line $L$ through $P$ that is not tangent to $X$ and doesn't go through a singularity of $X$ will cut $X$ in $d$ distinct points: $X\cap L=\\{Q_1,\cdots,Q_d\\}$.
If $X$ is not a line, $d\geq 2$ so that the the secant line $\overline {Q_1Q_2}$ is equal to $L$.
Thus every point $P$ is indeed on a secant of $X$, namely $\overline {Q_1Q_2}=L$, and this means that the secant variety of $X$ is indeed $\mathbb P^2$.