Prove that $\lim x_k = a \iff \lim \langle x_k, y \rangle = \langle a, y \rangle$ $\forall$ $y \in \mathbb{R}^n$. Could someone help me prove this real analysis theorem?

Prove that $\lim x_k = a \iff \lim \langle x_k, y \rangle = \langle a, y \rangle$ $\forall$ $y \in \mathbb{R}^n$.
Hint: For the first implication use the fact that the inner product is a continuous function. For the converse, let $y^i$ be zero in all entries and one in the $i_{\mathrm{th}}$ entry then you get $\lim\limits_{k\rightarrow\infty} &lt;x,y^i&gt;=\lim\limits_{k\rightarrow\infty} x^k_i=a_i$ where $x^k_i$ is the $i_{\mathrm{th}}$ entry of the vector $x^k$.
Hint It suffices to prove that 

$$\lim x_k - a=0 \iff \lim \langle x_k-a, y \rangle = 0 \,;\, y \in \mathbb{R}^n$$

$\Rightarrow$ follows immediately from $\left| \langle x_k-a, y \rangle \right| \leq \| x_k-a \| \| y\| \,.$

$\Leftarrow$ Write $ x_k-a=y_k u_k$ where $y_k \in [0, \infty)$ and $u_k$ is an unit vector.

Now $u_k$ lives in a compact space. It suffices to prove that the only cluster point of $y_k$ is $0$.

To do this, pick $t$ is a cluster point of $y_k$, pick $y_{k_n} \to t$ and pick a subsequent of $y_{k_n}$ for which $u_{k_n}$ converges to some $u$. Set $y=u$ in 

$$\lim \langle x_k-a, y \rangle = 0 $$