Suppose that $X(t)$ is a stochstic process with stationary and independent increments. Suppose also that $EX(1) = 0$, $EX^2(1) = 1$, and $X(t) \overset{d}{=} \sqrt{t} X(1)$, where $\overset{d}{=}$ means that both random variables have the same distribution.
Show that $ X(t) - X(s) \overset{d}{=} \mathcal{N}(0,|t-s|)$
I think that it is possible to use the Central Limit Theorem, but I am not sure how.
Suppose $t<s.$ If we then show this distribution we seek is $\mathcal N(0,s-t),$ then the same reasoning says that if $s<t$ then it is $\mathcal N(0,t-s),$ so either way it's $\mathcal N(0,|t-s|).$
With $t<s,$ partition the interval $[t,s]$ as
$$t = u_0 < u_1 < u_2 < \cdots < u_{n-1} < u_n = s $$
with equal increments $u_{i+1} - u_i.$ Then
\begin{align} & X(t) - X(s) \\[8pt] = {} & \big(X(u_1) - X(u_0)\big) + \big(X(u_2)-X(u_1)\big) + \cdots \\[6pt] & \qquad \cdots + \big(X(u_n) - X(u_{n-1}) \big) \end{align} This is a sum of a large number of i.i.d. terms with finite variance, so the central limit theorem applies.
Now suppose it were thought that the distribution is some particular distribution very close to a normal distribution but not exactly normal. In that case, use a value of $n$ so large that you'd have to be even closer to the normal distribution than, that, so that doesn't happen. Therefore it must be exactly normal.