I have to prove that $|T \cup S|$ where $T$ is infinite and $S$ is countable, equal to $|T|$, and this is also $|\mathbb{R}|$.
How can I approach this?
$|T \cup S| = |T| = |\mathbb{R}|$
I tried to approach it by considering the fact that $|S|$ is $N_{0}$. But after that I couldn't advance much.
You can't prove what you would like to prove. But let's see what we can say given the information.
We will assume that we have $AC$ so that sizes of sets are well behaved. Without $AC$ this would get a lot trickier.
We can then see that $|A\cup B|$ for any two set $A,B$ such that at least one is infinite is just $\max\{|A|,|B|\}$. This can be shown using an easy construction where for every odd ordinal we take an element of $A$ and for every even ordinal we take an element of $B$. If we run out of elements in one of the sets then we can just choose one we allready chose and that gives us a function that's not one-to-one but onto. That's enough since we get an upper bound on the size and the lower bound will be given by the size of the bigger inifinite set. The two will then coincide.
Once we know that the size of a union of two sets at least one of which is infinite is just max it's not hard to show that in your case $|S\cup T|=|T|$. Since both of the sets are inifinite and $|S|=\aleph_0=|\mathbb{N}|$ which is the smallest possible infinite set then the size of the union must be at least as large as $|T|$ since $|T|\geq \aleph_0$. That last claim is equivalent under AC to claiming that $T$ is inifinite.
So now to explain why you can't get the result you want. Well for one $|T|$ could just be $\aleph_0$. I.e. $T$ could just be a countable set itself in which case $|S\cup T|=\aleph_0$. You could also have $T=\mathcal{P}(\mathbb{R})$ in which case $|T|>|\mathbb{R}|$. Last but not least if you are working in a model where $\neg CH$ holds you might have a bunch of sizes between $\aleph_0$ and $|\mathbb{R}|=\mathfrak{c}$ and $|T|$ could be any one of those.