Suppose $X \subseteq \lbrace 0 , 1 \rbrace ^{m}$ such that $|X| \geq 2^{0.8m}$, and $m \geq 2$, then prove that there exists $x,y \in X$ with $||x - y||_{1} \geq m/2$.
My approach to prove this was if there is no such $x,y$, then $X$ is inside a Hamming ball of radius $m/2$ . But this does not give me a tight enough inequality on the size of $X$.
The problem of obtaining the largest set $X$ under the condition that its largest pairwise distance is bounded from above is known as the anticode problem, since it is naturally dual to the problem of finding an optimal error correcting code, where the smallest pairwise distance is bounded from below. It is apparently a theorem of Kleitman (http://dx.doi.org/10.1016/S0021-9800(66)80027-3) that the best binary anticode of diameter $2k$ and length $m$ is simply constructed by taking all strings of length $m$ with at most $k$ 1s (that is, a Hamming ball of radius $k$). If you let $k=\lfloor(\lceil m/2\rceil-1)/2\rfloor$ (I hope I got this right now), then you will find that the size of this anticode exceeds the bound you give in your question, and so the statement you cited is false, as already noted by Robert and Christian. It can be corrected by increasing $0.8$ to a slightly higher number, as they note.
Some more on Hamming anticodes: http://dx.doi.org/10.1006/aama.1998.0588
For small $m$, I've computed the cardinality $s(m)$ of the largest subset of $\{0,1\}^m$ of Hamming diameter $< m/2$, as follows:
$$\matrix{m & s(m)\cr 1 & 1\cr 2 & 1\cr 3 & 2\cr 4 & 2\cr 5 & 6\cr 6 & 7\cr 7 & 14\cr 8 & 16\cr}$$ The sequence does not appear to be in the OEIS. Clearly this is much too small for asymptotics (and with the methods I'm using it's going to be hard to get results for much larger $m$), but some of the sets returned seem to have a rather interesting structure. For example, for $m=8$ here is a set of size $16$ that has diameter $3$:
$$ \matrix{00000001\cr 00000010\cr 00000011\cr 00000101\cr 00000110\cr 00000111\cr 00001011\cr 00001111\cr 00010011\cr 00010111\cr 00100011\cr 00100111\cr 01000011\cr 01000111\cr 10000011\cr 10000111\cr }$$
EDIT: One lower bound of $s(m)$ is the cardinality of a ball of radius $\lfloor (m-1)/4 \rfloor$. This is $2^{m} \mathbb P(S_m \ge m - \lfloor (m-1)/4 \rfloor)$ where $S_m$ is the number of heads in $m$ independent fair coin flips. By Cramér's theorem from Large Deviations theory, as $m \to \infty$ this is roughly $ 2^m \exp(-m I(3/4)) = 2^{-km}$ where $k = 2 - (3/4) \log_2(3) \approx 0.811278$. Thus if $m$ is sufficiently large, a ball of radius $\lfloor (m-1)/4 \rfloor$ provides a counterexample.
Numerically, $377$ seems to be the least $m$ for which this works. That is, a ball of radius $94$ in $\{0,1\}^{377}$ has diameter $188 < 377/2$ and cardinality approximately $6.315655200 \times 10^{90} $, which is more than $2^{0.8 \times 377} \approx 6.175138852 \times 10^{90}$.
Just to show this isn't an artifact of roundoff error, the precise cardinality in this case is $$ \sum_{i=0}^{94} {377 \choose i} = 6315655199547126133494801496762823097854137171340937564314538960656350969227380734836894104$$
