Subsequence of a sequence 
If $\lim_{n\rightarrow\infty}{\langle a_n \rangle} = a$ and $\langle a_{in} \rangle$ is any subsequence of $\langle a_n \rangle$, then $\lim_{n\rightarrow\infty}{\langle a_{in} \rangle} = a$, but the opposite is not (necessarily) true.

I am trying to understand the above theorem. However, I am struggling to come up with any examples that prove this theorem and example the prove that the opposite to this theorem is not true. Are there any clear examples that show this theorem.

To show that the opposite is not necessarily true, consider the sequence that goes $$0,1,0,1,0,1,0,1,\ldots$$ This does not converge, as it fluctuates between $0$ and $1$ forever, but if I take the first term, the third term, the fifth term, and so on, I get the subsequence $0,0,0,0,\ldots$, which is just the constant sequence that converges to $0$.
For the other direction (that if a sequence is convergent, any subsequence will also converge) you can just pick up any convergent sequence $\{a_n\}$, and the theorem will be true for $\{a_n\}$. For instance, the sequence $\{\frac{1}{n}\}$ converges to $0$ as $n\to \infty$. Its terms are
$$1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots$$
Picking any subsequence of these terms will again give rise to a convergent sequence, with the same limit as the original. For instance, I can pick the subsequence
$$1,\frac{1}{2},\frac{1}{3},\frac{1}{5},\frac{1}{7},\frac{1}{11},\frac{1}{13},\frac{1}{17},\ldots$$
and it will be a sequence that has the limit $0$.
