Let $(a_n)^\infty_{n=0}$ be a sequence of real numbers, and let $L$ be a real number. Then the following two statements are logically equivalent (each one implies the other):
(a) The sequence $(a_n)^\infty_{n=0}$ converges to $L$.
(b) Every subsequence of $(a_n)^\infty_{n=0}$ converges to $L$.
By the definition of convergence if $(a_n)^\infty_{n=0}$ conveverges to $L$, then for every positive real $\epsilon$ there exist $N > 0$ such that $|a_n-L| < \epsilon$ for all $n > N$. Now take any subsequnce $(b_n)^\infty_{n=0}$ of $(a_n)^\infty_{n=0}$. Since $(b_n)^\infty_{n=0}$ is a subsequence then by definition there exist an increasing function $f: N \to N$, such that $b_n = a_{f(n)}$.
From (a to b):
If $f(n) > N$ for the given $N$ then by definition $|a_{f(n)} – L| = |b_n – L|< \epsilon$, so the subsequence converges to $L$. However, if $f(n) < N$ for all $n$, for example, $f(n) = 0$ for all $n$ and the subsequence is just {$a_0,a_0,a_0…$}, then what?
Suppose $b_n$ does not converge to $L$. Then there would exist an $\epsilon > 0$ such that for all $N > 0$ there exist an $n > N$ for which $|b_n – L| > \epsilon$ i.e. there exist an $f(n) > N$ for which $|a_{f(n)} – L| > \epsilon$. Which is a contradiction, isn't it?
From (b to a):
If every subsequence converges to $L$, then for any increasing $f:N \to N$ $b_n = a_{f(n)}$ converges to $L$. In particular take $f(n) = n$. Then $b_n = a_n$ converges to $L$.
Best Answer
Note that for any increasing function $f : \mathbb{N} \to \mathbb{N}$, we have $f(n) \geq n$ for all $n \in \mathbb{N}$ (why?). To prove (a) $\Rightarrow$ (b), since $a_n \to L$, we have $\forall \epsilon > 0$, there exists $M \in \mathbb{N}$ such that $|a_n - L| < \epsilon$ for all $n \geq M$. Since $f(n) \geq n \geq M$, $|a_{f(n)} - L| < \epsilon$ for all $n \geq M$ as well, so the subsequence must converge to the same limit.