Show that a convergent sequence in $\mathbb{R}^n$ has a unique limit in $\mathbb{R}^n$. I need help with this problem:

Show that a convergent sequence in $\mathbb{R}^n$ has a unique limit in $\mathbb{R}^n$.

So, I tried using a theorem that says that $(a_k)$ is a sequence in $\mathbb{R}^n$ with $a_k=(a_{k1},...,a_{ki})\ \forall k\in \mathbb{N}$, and $a=(a_1,...,a_n)\in \mathbb{R}^n$. Then $a_k\rightarrow a\iff a_{ki}\rightarrow a_i$ for each $i=1,...,n$.
I tried to show it by assuming that the statement is true when $n=1$ because my teacher said my could. Then If $(a_k)$ is a convergent series that means that $a_k\rightarrow a$, thus $a_{ki}\rightarrow a_i$. Since it is convergent and has a unique limit in $\mathbb{R}$, $a_{k1}\rightarrow a_1$ and $a_{k2}\rightarrow a_2\Rightarrow \ a_1=a_2$. Thus it works $\forall i=1,...,n$. Is my proof correct? If not, how can I fix it? Thanks.
 A: There is one error. 
$a_{k1} \to a_1$ and $a_{k2} \to a_2$ does not imply $a_1=a_2$, because $a_1$ and $a_2$ are different coordinates of the limit.
To make your proof complete, I would add:

Since the limits of the sequences $a_{ki}$ (namely $a_i$) are unique, and since $a=(a_1,a_2,...,a_n)$, we have that the limit $a$ is unique as well. This follows from the fact that two ordered $n-$tuples are equal iff their respective coordinates are equal.

To elaborate more, consider the following.
Suppose for a contradiction that $b$ is another limit of the sequence $a_k$, and $b\neq a$. This means that $b_i \neq a_i$ for some $i$. But, as each coordinate of the limit is unique (based on the assumption your teacher permitted for $n=1$), having $b_i \neq a_i$ means that $b_i$ is not the limit of the sequence $a_{ki}$. This is a contradiction. Thus, we conclude that $b=a$.

Although this is a valid proof, the one that @Dave suggested is much more elegant:
Suppose that $a$ and $b$ are two different limits of the sequence.
We have that $$0\leq\Vert a-b\Vert \leq \Vert a-a_k \Vert + \Vert a_k-b \Vert \to 0$$
By the squeeze theorem we have that $$\lim\limits_{k\to\infty} \Vert a-b \Vert = 0$$
Because $a$ and $b$ are constants, this implies that $a=b$.
If @Dave decides to post this as his own answer, I will remove this alternate proof from my answer.
