Since you have a finite-dimensional inner product space, you can always find an orthonormal basis (via Gram-Schmidt). Let $(e_1, \ldots, e_n)$ be such a basis for an inner product space $X$, and suppose $K \subseteq X$ is closed and bounded by $M$ (i.e. $v \in K \implies \|v\| \le M$).
Consider the linear maps $\langle \cdot, e_i \rangle$. Using Cauchy-Schwarz, you can show that these linear maps are continuous. Moreover, by the same inequality, if $x \in K$, then
$$|\langle x, e_i \rangle| \le \|x\| \cdot \|e_i\| = \|x\| \le M.$$
Consider a sequence $(x_i) \in K$. The sequence $\langle x_i, e_1 \rangle$ is contained in the compact interval $[-M, M]$, and so there must be a convergent subsequence, say, $(x_{i_j})$. Then $\langle x_{i_j}, e_2\rangle$ is a sequence contained in $[-M, M]$, so another convergent (sub-)subsequence. A third (sub-sub-)subsequence guarantees that $\langle x_i, e_3\rangle$ converges, etc. After $n$ iterations (a sub-sub-...-subsequence), you'll have a sequence (call it $(y_j)$) such that $\langle y_j, e_i \rangle$ converges to some $a_i \in [-M, M]$ for all $i = 1, \ldots, n$ (as $j \to \infty$). This sequence will be a subsequence of $(x_i)$.
So, with this in mind, let's prove that $y_j \to a_1 e_1 + \ldots + a_n e_n =: y$. By a standard result, we have, for each $j$,
$$y_j = \langle y_j, e_1 \rangle e_1 + \ldots + \langle y_j, e_n \rangle e_n.$$
Thus, by Pythagoras's theorem,
$$\|y_j - y\|^2 = |\langle y_j, e_1 \rangle - a_1|^2 + \ldots + |\langle y_j, e_n \rangle - a_n|^2,$$
which is the sum of $n$ sequences that all converge to $0$, hence $\|y_j - y\| \to 0$, i.e $y_j \to y$. That is, we started with an arbitrary sequence $(x_i)$, formed a subsequence $(y_j)$, and constructed a limit for the sequence. How do we know that $y \in K$? Because $K$ is closed!