I think the most helpful thing you can do here is to think about what decimal expansions really are.
What do we mean when we write $x=1.234999\ldots$? As you probably know, finite decimal expansions are just sums of certain decimal fractions, for example: $1.234$ means is by definition $1+\frac2{10}+\frac3{100}+\frac4{1000}$. So infinite decimal expansions are in fact just another way to write an infinite series, in your case: $$1.234999\ldots = 1+\frac2{10}+\frac3{10^2}+\frac4{10^3}+\frac9{10^4}+\frac9{10^5}+\frac9{10^6}+\cdots$$ which we may write more concisely as: $$\pm a_0.a_1a_2a_3a_4\ldots = \pm\sum_{n=0}^\infty\frac{a_n}{10^n},$$
where $a_0\in{\mathbb N_0}$ and $a_i\in\lbrace0,1,2,3,4,5,6,7,8,9\rbrace$ for $i\in\mathbb N$.
In your case, $a_0=1,a_1=2,a_2=3,a_3=4$ and $a_i=9$ for all $i>3$. So we are going to solve the problem if we determine the value of this series: $$1.234999\ldots =1+\frac2{10}+\frac3{10^2}+\frac4{10^3}+\frac9{10^4}+\frac9{10^5}+\frac9{10^6}+\cdots =\\=1.234+\frac9{10^4}(1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\cdots).$$ Now the expression in parentheses is a geometric series, i.e. a series of the form $1+x+x^2+x^3+\cdots$ (or if you prefer the more concise notation, $\sum\limits_{n=0}^\infty x^n)$. As you probably already know (otherwise you will most likely learn this soon), the geometric series converges to $\frac1{1-x}$ for $-1<x<1$. So in our case where $x=\frac1{10}$, we have: $$\sum_{n=0}^\infty(\frac1{10})^n =1+\frac1{10}+\frac1{10^2}+\frac1{10^3}+\cdots = \frac1{1-\frac1{10}}=\frac{10}9.$$ Now we just plug this into the expression above and we get $$1.234999\ldots = 1.234 + \frac9{10^4}\frac{10}9 = 1.234+\frac1{10^3} = 1.234+0.001=1.235$$ Thus the two numbers are equal.
To see that for two distinct real numbers $a,b$ we indeed have a rational between them, we note that $c=\frac{a+b}2$ is a number strictly between them. But as every rational number has a decimal expansion, we get a sequence of truncated decimal expansions, $\sum\limits_{n=0}^N\frac{c_n}{10^n} = c_0+\frac{c_1}{10}+\frac{c_2}{10^2}+\cdots+\frac{c_N}{10^N}$, that converges (gets arbitrarily close) to this number. These are rational numbers. Now, as $a$ and $b$ are both a positive distance away from $c$, this means there will be a number in this sequence that is closer to $c$ than $a$ and $b$ are. This means we have found a rational number strictly between $a$ and $b$, so we are done.
Added: The thing to remember here is that some real numbers have two distinct decimal expansions. If you think about it a bit, you will see that these are precisely the numbers that have a terminating decimal expansion, i.e. the numbers which have $0$ from some place on. (These are the rational numbers which can be written in the form $\frac{m}{10^k}$ for some $m\in\mathbb Z$ and $k\in\mathbb N_0$.) These numbers can be written in two ways because we can always replace the last digit before the zeroes with a digit that is one smaller and add nines after it. Note that a funny thing happens with $0$. It may still be regarded as having two decimal expansions in a sense: $0=+0.000\ldots=-0.000\ldots$ This is because we have defined the decimal expansion using the formula $$\pm a_0.a_1a_2a_3a_4\ldots = \pm\sum_{n=0}^\infty\frac{a_n}{10^n},$$
for $a_0\in{\mathbb N_0}$ and $a_i\in\lbrace0,1,2,3,4,5,6,7,8,9\rbrace$ for $i\in\mathbb N$. The strange behaviour of $0$ arises because this way we treat positive and negative numbers differently. It goes away if we use a different formula: $$a_0.a_1a_2a_3a_4\ldots = \sum_{n=0}^\infty\frac{a_n}{10^n}$$ where $a_0\in{\mathbb Z}$ and $a_i\in\lbrace0,1,2,3,4,5,6,7,8,9\rbrace$ for $i\in\mathbb N$. I don't believe this is completely standard, however, probably because in this case the other decimal expansion looks a bit funny: $(-1).999\ldots=0.000\ldots=0$. Note, though, that using this (possibly impractical) notation every number that can be written as $\frac{m}{10^k}$ behaves the same. In both cases every number that is not of this form has a unique decimal expansion.
