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Can someone please verify if my proof is correct, and add to it if it isn't?

Since we're given that $B$ is countable, we know $B$ is either finite or has the same cardinality as, say, the set $mathbbQ$. So $vert B vert = aleph_0$.

Similarly, if $A$ is infinite, then we know the result $aleph_0 leq vert A vert$.

This implies that $ vert B vert leq vert A vert$.

Then $vert A cup B vert = vert A vert$.

Am I missing something here?

I am not at all expert in large cardinals, but as far as my understanding goes, once you have proved $|B| = aleph_0$, you then have (assuming that $A$ and $B$ are disjoint) $|Acup B| = |A| + |B| = |A| + aleph_0$, and the arithmetic cardinals is such that adding $aleph_0$ to any infinite cardinal does not change said cardinal.

That being said, this "arithmetic" property is no less than what you are asked to show here, and I think the point of the question is to make you prove that. As @5xum suggested in his comment, you can define an injection $A to Acup B$ together with a surjection $Ato Acup B$ and use the Cantor-Bernstein theorem.

To get a feel for how to do that, I would suggest to try to understand how a bijection between $mathbbN$ and $mathbbN^2$ works.
Also note that without loss of generality, you can assume $B = mathbbQ$ ou $mathbbN$. Indeed since $|mathbbQ| = |mathbbN| = aleph_0$, you have bijections $BtomathbbN$ and $BtomathbbQ$. So if you get a bijection $Ato Acup mathbbN$, you can then use the fact that the composition of bijections is a bijection.
Finally you can assume that $mathbbNsubset A$ (more precisely, there is a subset of $A$ which is in bijection with $mathbbN$). This should help you mimic the construction of the case $mathbbN^2$

You are missing something, yes. You only wrote down the following relations:

• 
  $|B|=aleph_0$, a relation that is not even true! There is nothing in the question that prevents $B$ from being a finite set.


• $|B|leq |A|$


• $aleph_0leq |A|$

There is nothing here for you to conclude that $|Acup B|=|A|$. For that, you would have to prove that $|A|leq |Acup B|$ (this should be easy) and that $|Acup B|leq |A|$ (this one should prove a little harder, as you need a surjection from $A$ to $Acup B$ to prove this inequality).