The equation #x = e^-x# has a real root in #[0,1]# if and only if the function

#f(x) = e^-x - x#

has a zero in #[0,1]#.

#f(0) = e^(-0) - 0 = 1#

and

#f(1) = e^(-1) - 1 = 1/e - 1 < 1 - 1 = 0#,

so #f(x)# undergoes at least one sign change in #[0,1]#. This is a consequence of the continuity of #f(x)# and the Intermediate value theorem , which in this case says the following:

Since #f(0)>0# and #f(1) < 0#, there exists some real number #x_m# in #(0,1)# such that #f(x_m) = 0#, and we have proved the existence of a real zero of #f(x)# in #[0,1]#.

Due to the relation between the zeros of #f(x)# and the roots of #x = e^-x#, we can conclude that #x = e^-x# has a real root in #[0,1]#.

EDIT: As pointed out by George, the function #f(x)# must be continuous and defined on the *entire* interval of interest. You can, as an exercise, show that this is in fact the case. One shortcut is to use that #f(x)# is a sum of two analytic functions in combination with (https://proofwiki.org/wiki/Combination_Theorem_for_Continuous_Functions/Sum_Rule) .