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I started the proof, but couldn't figure out where to go... here is my attempt:

Take $S, T \in B(X)$, the space of operators from $X$ to $X$, such that $ST(x) = TS(x)$ for all $x \in X$. Then take $y \in \ker(T)$ so that $T(y) = 0$. Then, $S(T(y)) = S(0)$. However, since $T$ and $S$ commute we have that $$T(S(y)) = S(T(y)) = S(0)$$ as well.

I might be missing something foundational here, but $S(0)$ doesn't necessarily have to equal $0$ as an operator?

Then, as far as the closure of the image space of $T$, I'm not really sure either. Thank you.

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    $S\in B(X)$ means $S(0)=0$. Furthermore, what exactly do you mean by "preserves" those subspaces? Do you mean they are invariant for $S$? – 2017-01-19
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    @Aweygan Yes, invariant. Also, $S \in B(X)$ then means $B(X)$ is actually the space of linear operators from $X$ to $X$? – 2017-01-20
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    Yes, that's relatively standard – 2017-01-20
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    @Aweygan Okay, that clears that up fairly easily then, thank you. – 2017-01-20
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    Yep. And another note from reading the comments shared with Omnomnomnom, your functional analysis book will probably use "operator on $X$" to denote "bounded linear map from $X$ to $X$", at least until a possible chapter on unbounded operators. – 2017-01-20

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$x\in Ker T, T(x)=0$, $T(S(x))=S(T(x))=0$, $S(x)\in Ker T$.

$y$ in the adherence of $Im(T)$, $y=lim_nT(x_n)$, $S(y)=S(lim_nT(x_n))=(lim_nST(x_n))$ since $S$ is bounded, so $S(y)=lim_nST(x_n)=lim_nT(S(x_n))$ in the adherence of $Im T$.

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    You can use `\operatorname{Ker}(T)` and `\operatorname{Im}(T)` to get $\operatorname{Ker}(T)$ and $\operatorname{Im}(T)$. As for limit, `\lim` will suffice. – 2017-01-19
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The $0$ in $S(0)$, as you've written it, is the $0$-vector (which was $T(y)$). For any linear operator $S$, $S(0) = 0$. That is, every linear operator takes the $0$-vector to the zero vector.

As for the image: note that $x$ is in the image of $T$ if $x = T(y)$ for some vector $y$. However, in this case: $$ S(x) = S(T(y)) = T(S(y)) = T(\text{[a vector]}) $$ which means that $S(x)$ is in the image of $T$. Now, use continuity to "carry this over" to the closure.

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    The way my text reads is that $B(X)$ is the space of operators from $X$ to $X$; I'm new to the notation though, is that actually the space of linear operators from $X$ to $X$ then? – 2017-01-20
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    Actually, $B(X)$ usually means the space of **bounded** linear operators. But yes: these are linear operators, at the very least. Otherwise, it wouldn't be in a functional analysis textbook. – 2017-01-20