We want to show that $$ {\rm P}[X \ge {\rm E}(X|U) - 1/m] \ge {\rm P}(U)/m. $$ This is trivially not true if $0 < m < 1$, for if $U = \Omega$, the right-hand side is greater than $1$. So, assume that $m \geq 1$, and ${\rm P}(U)>0$. Further, if ${\rm E}(X|U) - 1/m \leq 0$, then the inequality is trivial; hence we assume that ${\rm E}(X|U) - 1/m > 0$. Let $P_{X|U}$ denote the conditional distribution of $X$ given $U$, so that $$ P_{X|U} {(A)} = {\rm P}(X \in A | U) = \frac{{{\rm P}([X \in A],U)}}{{{\rm P}(U)}}. $$ Then, $$ {\rm E}(X|U) = \int_{[0,1]} {xP_{X|U} ({\rm d}x)}. $$ Next, decompose $[0,1]$ into the union of the intervals $I_1 = \lbrace 0 \leq x < {\rm E}(X|U) - 1/m \rbrace$ and $I_2 = \lbrace {\rm E}(X|U) - 1/m \leq x \leq 1 \rbrace$. On the one hand, $$ \int_{I_1 } {xP_{X|U} ({\rm d}x)} \le [{\rm E}(X|U) - 1/m]P_{X|U} (I_1 ) \le {\rm E}(X|U) - 1/m, $$ and on the other hand, $$ \int_{I_2 } {xP_{X|U} ({\rm d}x)} \leq \int_{I_2 } {P_{X|U} ({\rm d}x)} = P_{X|U} {(I_2)} \leq \frac{{{\rm P}(X \in I_2 )}}{{{\rm P}(U)}}. $$ Combining it all, we get $$ {\rm E}(X|U) \leq {\rm E}(X|U) - 1/m + {\rm P}(X \in I_2)/{\rm P}(U). $$ That is ${\rm P}(X \in I_2) \geq {\rm P}(U)/m$, which is exactly what we want.