Your set has measure zero by theorems of Khinchin. First a theorem of Khinchin shows that for almost all real numbers $x$ (i.e. outside a set of measure zero) one has
$$
\lim_{n\to \infty} \frac{\log q_n}{n}= C
$$
for a positive constant $C$. Here $p_n/q_n$ are the convergents of $x$. So almost surely, the denominators $q_n$ are exponential in $n$. In fact the constant $C$ here is known to be $\pi^2/(12 \log 2)$ -- the Levy or Khinchin-Levy constant.

Next another Theorem of Khinchin shows that for almost all $x$ one has infinitely many approximations $p/q$ with
$$
\Big| x-\frac{p}{q}\Big| \le \frac{1}{q^2\log q}.
$$
Here what is used is that $\sum_q 1/(q\log q)$ diverges.

Putting both results together (and since continued fractions give best rational approximations) we see that for almost all $x$ we have infinitely many $n$ such that
$$
\Big| x-\frac{p_n}{q_n} \Big | \le \frac{C}{n q_n^2},
$$
for some positive constant $C$. Now recall that for any $x$ we have $|x-p_n/q_n|$ is about size $1/(q_nq_{n+1})$ and $q_{n+1}$ is about size $a_{n+1}q_n$. Therefore it follows from the above inequality that for infinitely many $n$ one has $a_n \ge cn$ for some constant $c>0$.

**Edit**: I just got the chance to look at Khinchin's book Continued Fractions, and he discusses this question explicitly in Theorem 30 of that book (page 63 in the Dover edition). Let $\phi(n)$ be an arbitrary positive function ($n\in {\Bbb N}$). Then the inequality $a_n \ge \phi(n)$ is satisfied for infinitely many $n$ for almost all real numbers $x$, provided $\sum 1/\phi(n)$ diverges. On the other hand if $\sum 1/\phi(n)$ converges, then this inequality is satisfied for only finitely many $n$ (and almost all $x$).