I want to show that the following is a closed-compact-Lie subgroup of $SU(2)$ $$T=\left\{\begin{pmatrix}e^{i\phi}&0\\0&e^{-i\phi}\end{pmatrix} : \phi \in \mathbb{R}\right\}.$$ By Cartans closed subgroup theorem and since $SU(2)$ is compact it is enough to show that $T$ is a closed.
I think I am getting confused on the the idea of something being closed in topology. Initially I thought that a set is closed if it contains all its limit points. That is for any sequence $(x_n)$ in $T$ with limit $x \in S$ we have to show that $x\in T$. (This seems strange, what does it mean for a sequence in $T$ or even in $SU(2)$ to have a limit)
Assuming the previous stuff is correct, I think we do this by taking a continous function $f:SU(2)\rightarrow SU(2)$ such that $f(x)=I$ for $x\in T$ and $f(x)\neq I$ for $x \notin T$. Then for for the sequence we get $\lim f(x_n)=\lim I = I \in T$ and by continuity we get $\lim f(x_n)=f(x)=I$ so $x\in T$.
But I cannot find such an $f$. Is this the correct approach?
There are several ways to describe the usual topology on $SU(2)$, but probably the simplest is that it's the subspace topology from the space $\mathbb{C}^{2 \times 2}$ of $2 \times 2$ complex matrices, which you may identify with $\mathbb{C}^4$. So, a sequence in $SU(2)$ converges iff it converges elementwise, i.e. we have $\left(\begin{smallmatrix} a_n & b_n \\ c_n & d_n \end{smallmatrix}\right) \to \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ iff we have $a_n \to a$, $b_n \to b$, $c_n \to c$, and $d_n \to d$ as sequences of complex numbers.
So suppose you have a sequence of elements $x_n$ from $T$. By definition of $T$, each $x_n$ must be of the form $\left(\begin{smallmatrix} e^{i \phi_n} & 0 \\ 0 & e^{-i\phi_n} \end{smallmatrix}\right)$ for some $\phi_n$. Now suppose that there is some $x = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)$ with $x_n \to x$. This means that $e^{i \phi_n} \to a$, $0 \to b$, $0 \to c$, and $e^{-i \phi_n} \to d$. You have to show that $b=c=0$ (trivial) and that there exists some $\phi$ such that $e^{i \phi} = a$, $e^{-i \phi} = d$.
A different approach is to find a homeomorphism from $T$ to the circle $S^1$, say the unit circle of $\mathbb{C}$, which hopefully you know is compact.