Let $a$ and $b$ be distinct irrationals; we lose no generality to suppose $a<b$. Assume they have equal integer parts, since otherwise there is an integer between them and the question is trivial. They have infinite decimal expansions, $.a_1a_2a_3\ldots$ and $.b_1b_2b_3\ldots$. These cannot agree in every position since otherwise $a=b$. So say they agree up to the $n-1$th position and differ at the $n$th. Then $$ x= \;.b_1b_2b_3\ldots b_n 0000000\ldots$$

is a rational number strictly between $a$ and $b$:

$$\begin{align}
a &= \;.a_1a_2\ldots a_{n-1}a_n\ldots \\& <\; .a_2a_2\ldots a_{n-1}b_n000\ldots &= x
\end{align}$$

because $a_n < b_n$, and

$$\begin{align}
x & = \;.b_1b_2b_3\ldots b_n 0000000\ldots \\ & < \;.b_1b_2b_3\ldots b_nb_{n+1}\ldots & = b
\end{align}$$

because not all of $b_{n+1}, b_{n+2}, \ldots$ can be zero.

For example, there is a rational number betwen $\sqrt2 = 1.4142\ldots$ and $\sqrt3 -\frac14 = 1.482\ldots$; this method produces the rational number $1.48000\ldots = \frac{37}{25}$. You can of course do the same thing in base 2; then you get $x = 1._20111000\ldots = \frac{23}{16}$, which also works.

It would also be fairly easy to produce a similar argument based on the continued fraction expansions of $a$ and $b$, but I think this is simpler.