I thought I understood the problem until I read the solution.
"Since the distribution function for $X$ is continuous (and differentiable) for $x>1$, it follows that $P[X=x]=0$ for $x>1$."
How can this be when it says that $P[X≤x]=1-e^{-x}$ for $x≥1$? Also, how does this eliminate choices $A$, $B$, and $C$?
When the solution refers to the "distribution function," it's referring to the cumulative distribution function; i.e., $$F_X(x) = \Pr[X \le x].$$ This is of course continuous for $x > 1$, so on this interval, the random variable $X$ doesn't consist of discrete probability masses, but a continuous probability density. If we differentiate on this interval, we get $$f_X(x) = e^{-x}, \quad x > 1,$$ but recall that $$\Pr[X = x] \ne f_X(x).$$ A density is not the same as a probability mass.
To illustrate, I could use the CDF to get for example $\Pr[1 < X < 2]$, but I could not use it to get $\Pr[X = 2]$, because, for example, $$\Pr[X = 2] = \Pr[X \le 2] - \Pr[X < 2] = (1 - e^{-2}) - \lim_{x \to 2^-} (1 - e^{-x}) = 0.$$
