Show that if $\left\Vert Px-Qx\right\Vert <\left\Vert x\right\Vert$ for every $0\ne x\in V$, then $\dim M=\dim N$ 
Let $V$ be an inner product finite space and let $M,N\subseteq V$ subspaces. Assume $P$ and $Q$ are the projections on $N$ and $M$, respectively.
Show that if $\left\Vert Px-Qx\right\Vert <\left\Vert x\right\Vert$ for every $0\ne x\in V$, then $\dim M=\dim N$.

My try:
Assuming WLOG $dimN<dimM$
Let $B_N$ and $B_M$ be basis' for $N$ and $M$ respectively. Hence there exist $b_k\in B_M$ s.t $b_k\notin N$
$P|_{M}(b_{k})=b_{k}$
$Q|_{N}(b_{k})=0$
But I'm stuck here.
 A: By the Gram-Schmidt process, there exist vectors $e_{1}, \dotsc, e_{d}$ of $V$ such that $\left( e_{1}, \dotsc e_{m} \right)$ is an orthonormal basis for $M$ and $\left(e_{1}, \dotsc, e_{d} \right)$ is an orthonormal basis for $V$, where $m = \dim(M)$ and $d = \dim(V)$.
For all $x \in V$, we have $$p_{M}(x) = \sum_{k = 1}^{m} \left\langle e_{k}, x \right\rangle e_{k} \, \text{,} \quad x = \sum_{k = 1}^{d} \left\langle e_{k}, x \right\rangle e_{k} \quad \text{and} \quad \lVert x \rVert^{2} = \sum_{k = 1}^{d} \left\langle e_{k}, x \right\rangle^{2} \, \text{.}$$
By hypothesis, for every $x \in N \setminus \lbrace 0 \rbrace$, we have $$\sum_{k = m +1}^{d} \left\langle e_{k}, x \right\rangle^{2} = \left\lVert -\sum_{k = m +1}^{d} \left\langle e_{k}, x \right\rangle e_{k} \right\rVert^{2} = \left\lVert p_{M}(x) -p_{N}(x) \right\rVert < \lVert x \rVert^{2} = \sum_{k = 1}^{d} \left\langle e_{k}, x \right\rangle^{2} \, \text{,}$$ that is $\sum_{k = 1}^{m} \left\langle e_{k}, x \right\rangle^{2} > 0$, and hence there exists $k \in \lbrace 1, \dotsc, m \rbrace$ such that $\left\langle e_{k}, x \right\rangle \neq 0$.
Therefore, the linear map $\varphi \colon N \rightarrow \mathbb{R}^{m}$ defined by $$\varphi(x) = \left( \left\langle e_{1}, x \right\rangle, \dotsc, \left\langle e_{m}, x \right\rangle \right)$$ is injective, and hence $\dim(N) \leq m = \dim(N)$.
P.S.: In order to avoid possible confusions, I denote by $p_{M}$ (respectively $p_{N}$) the orthogonal projection on $M$ (respectively $N$).
A: I think you're on the right track. Let $B_N$ and $B_M$ be the bases and let $n_i$, $m_i$ be the basis vectors. Define $m^N_i = Pm_i$, so this means we project all the $m_i$ onto $N$. Form some linear combination $v=\sum_i\alpha_im_i^N$. Since $\dim M>\dim N$, there are "too many" $m_i^N$, more precisely, we can pick $\alpha_i^*$ not all zero so that $v=0$. Now define $w=\sum_i\alpha_i^*m_i\in M$. Since $m_i$ form a basis, $w\neq0$. However, by construction, $Pw=v=0$ and $Qw=w$. We immediately get
$$
\Vert Pw-Qw\Vert=\Vert0-w\Vert=\Vert w\Vert
$$
But this is a contradiction, so $\dim M=\dim N$.  
EDIT: This proof can be condensed, thanks to v_lentin. (Check the comments.) The restriction $P\vert_M$ is a linear map from $M$ to $N$, thus it maps a higher-dimensional space into a lower-dimensional one and can't be injective. This time, define $w$ as any nonzero element of $\ker(P\vert_M)$. The proof proceeds as before.
