Convergent Sequence with Finite Number of Terms Deleted is Convergent

Theorem
Let $\left({X, d}\right)$ be a metric space.

Let $\left \langle {x_k} \right \rangle$ be a sequence in $X$.

If $\left \langle {x_k} \right \rangle$ converges, the deletion of a finite number of terms will still result in a convergent sequence.

Similarly, if $\left \langle {x_k} \right \rangle$ diverges, the deletion of a finite number of terms will still result in a divergent sequence.

Proof
Suppose, the sequence $\left \langle {x_k} \right \rangle$ converges to $x \in X$, i.e. for every $\varepsilon > 0$ there is some index $N$ such that $d(x_n, x) < \varepsilon$ for $n \geq N$. The same $N$ will work for the new sequence with finitely many terms removed, so the new sequence converges to the same point $x$ as the original sequence.

For the second part note that also adding finitely many terms to a convergent sequence will still result in a convergent sequence. This implies that removing finitely many terms from a divergent sequence will still result in a divergent sequence (if it were convergent then the original sequence must already have been convergent).