Prove that if $\{v_1,v_2,\dots,v_n\}$ is orthonormal, then for any $v$, $\lVert v\rVert^2 \ge \Sigma_{j=1}^n \lvert \langle v,v_j \rangle \rvert$ 
Prove that if $S=\{v_1,v_2,\dots,v_n\}$ is an orthonormal set in $V$, then for any $v \in V$, $\lVert v\rVert^2 \ge \Sigma_{j=1}^n \lvert \langle v,v_j \rangle \rvert$. Also, prove that equality only holds if $dim(V)=n$.

I think there are two cases, either $v \in span(S)$ or $v \in span(S)^\perp$.  We know that if $v \in span(S)^\perp$, then $\Sigma_{j=1}^n \lvert \langle v,v_j \rangle \rvert=0$  However, then if $v \in span(S)$, then $v$ can be written as $v=\Sigma_{i=1}^n\frac{\lvert \langle v,v_i\rvert \rangle}{\lVert v_i \rVert}v_i$. 
I think all that I've stated above is correct, but now I need to synthesize the result from $v$ being in the span of $S$, which I'm stuck at.

It is not necessarily true that a vector belongs to a subspace or to its orthogonal. Example: $E=\{x,0):\ x\in\mathbb R\}$ is a subspace of $\mathbb R^2$, and $(1,1)$ is neither in $E$ nor its orthogonal. 
The argument you need is actually simple. Expand $v_1,\ldots,v_n$ to an orthonormal basis $v_1,\ldots,v_n,w_1,\ldots,w_m$. Then, for any $v$, we have 
$$
v=\sum_{j=1}^n\alpha_j v_j+\sum_{k=1}^m\beta_k w_k. 
$$
So
$$
\|v\|^2=\sum_{j=1}^n|\alpha_j|^2+\sum_{k=1}^m|\beta_k|^2
=\sum_{j=1}^n|\langle v,v_j\rangle|^2+\sum_{k=1}^m|\beta_k|^2. 
$$
Now conclude. 
