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I try to prove that in MSO we can't define equality of two cardinalities of structure. For example number of nodes in two graphs can't be expressed.

My only tool to do this is theorem: 

Language is regular iff it can be defined in MSO

So look please at my trial and try to check it:
We know that language of words such that number of symbol $a$ is equal to number of symbol $b$ is irregular. We get arbitrary word and construct two graphs - both without edges - only requirement is that number of $a$ is number of nodes and number of $b$ is equal to number in second graph.

Then we assume that there exists formula stated in content of task and we are able to define irregular language. Contradiction.

I know that it is not formal, but I can't do it better.

Edit
After suggestion answerer:
$$\phi(X,Y) \leftrightarrow|X|=|Y|$$ $$\exists_X\exists_Y[\phi(X,Y) \wedge (\forall_x (P_a(x)\to x\in X)) \wedge (\forall_x (P_b(y)\to y\in Y)\wedge (\forall_x (x\in X \vee x\in Y))] $$

$P_a(x)$ means that at position $x$ there is a symbol $a$.
I managed (I believe) define language with equal number of symbol $a$ and symbol $b$.

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    I think both of your $\to$ should be $\leftrightarrow$, and one of your $\forall x$ should be $\forall y$, but otherwise this looks fine to me. – 2017-02-17

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Sounds like the right direction to go in.

To make it less handvawy, you should explain exactly how, given a formula $\varphi(X,Y)$ which expresses that $X$ and $Y$ are equinumerous sets, you would construct a formula that recognizes the language $\{\mathtt a^n\mathtt b^n\mid n\in\mathbb N\}$.

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    I edited. Look again, please :) – 2017-02-17
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    @HaskellFun: I don't see any edit. – 2017-02-17
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    You should it now, @Henning Makholm – 2017-02-17