Proving that a group is closed under multiplication by scalars I need to prove that the group $U = \{A \in M_n(\mathbb{R})\mid AB=BA\}$ is closed under multiplication by scalars.

So I let $\alpha \in \mathbb{R}$ and let $A \in U$, which means I need to prove that $\alpha A \in U$.

$(\alpha A)B = \alpha(AB) = \alpha(BA)$

But that does not prove what I need, I need to prove that $B(\alpha A) \in U$.

Where do I go from here?
It seems that the $U$ is a subspace of $V=M_{n\times n}(\mathbb R)$. So you are on a right path. In fact, if $A,C\in U$ then $$(aA+bC)=(aA)B+(bC)B=a(AB)+b(CB)=...=B(aA)+B(bC)=B(aA+bC)$$. Note that I amasuuming $B$ is a certain fixed matrix in vector space $V$
No, you are not asked to prove that $B(\alpha A)\in U$. What you should prove is the following:


  if $A\in U$, then $\alpha A\in U$ for any $\alpha\in\mathbb{R}$,


or equivalently, if $AB=BA$, then $(\alpha A)B=B(\alpha A)$ for any scalar $\alpha$. This should be trivial because both sides of the last equality are equal to $\alpha(AB)$.

Alternatively, if we define $f:A\mapsto AB-BA$, then $f$ is a linear map defined on $M_n(\mathbb{R})$ and $U$ is its nullspace. Since every vector subspace is closed under scalar multiplication, the assertion follows.