If two projections are close, then they are unitarily equivalent

The result holds in a C*-algebra. The sketch of the proof I know, which was given to me as a series of exercises in a class on C*-algebras, is as follows. First one constructs isomorphisms from each C*-algebra on the following list to the next one (below all C*-algebras are unital, $\mathbb{Z}_2$ denotes the cyclic group of order $2$, and $M_2$ denotes the C*-algebra of $2 \times 2$ complex matrices):

• The free C*-algebra on two projections $p, q$,
• The free C*-algebra on two self-adjoint unitary elements,
• The group C*-algebra $C^{\ast}(\mathbb{Z}_2 \ast \mathbb{Z}_2)$,
• The crossed product $C(\mathbb{T}) \rtimes_{\alpha} \mathbb{Z}_2$ where $\mathbb{Z}_2$ acts on $\mathbb{T}$ by complex conjugation,
• The C*-algebra $A$ of continuous functions $g : [0, 1] \to M_2$ such that $g(0)$ and $g(1)$ both have the form $\left[ \begin{array}{cc} \alpha & \beta \\ \beta & \alpha \end{array} \right]$ for some $\alpha, \beta$ (possibly different for $0$ and $1$).

Under one such chain of isomorphisms $p$ gets sent to the function

$$[0, 1] \ni x \mapsto \left[ \begin{array}{cc} \frac{1}{2} & \frac{e^{\pi i x}}{2} \\ \frac{e^{-\pi i x}}{2} & \frac{1}{2} \end{array} \right]$$

and $q$ gets sent to the function with constant value $\left[ \begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{array} \right]$.

The imposition of the additional constraint $\| p - q \| = \lambda < 1$ corresponds to working in a quotient of $A$ where we consider functions $g : [0, x_0] \to M_2$ where $\sin \frac{\pi x_0}{2} = \lambda < 1$ (this condition is precisely what is needed for the sup norm of $p - q$, as a function to $M_2$, to be equal to $\lambda$). Then it suffices to write down a unitary equivalence between $p$ and $q$ in this algebra, but it becomes straightforward to do this explicitly: in fact we can take

$$u : x \mapsto \left[ \begin{array}{cc} 1 & 0 \\ 0 & e^{\pi i x} \end{array} \right].$$

(Note that $u$ does not define an element of $A$.)

Edit: I'm leaving the old post below, but before I want to write the proof as suggested by Bruce from his book, which uses the ideas in a more efficient way.

Assume that $\|p-q\|<1$, with $p,q\in A$, a unital C$^*$-algebra. Let $x=pq+(1-p)(1-q)$. Then, as $2p-1$ is a unitary, $$ \|1-x\|=\|(2p-1)(p-q)\|=\|p-q\|<1. $$ So $x$ is invertible. Now let $x=uz$ be the polar decomposition, $z=(x^*x)^{1/2}\in A$. Then $u=xz^{-1}\in A$. Also, $px=pq=xq$, and $qx^*x=qpq$, so $qx^*x=x^*xq$, and then $qz=zq$. Then $$ pu=pxz^{-1}=xqz^{-1}=uzqz^{-1}=uqzz^{-1}=uq. $$ So $q=u^*pu$.

============================================= (the old post starts here)

(A good friend pointed me to the ideas in this answer, so I'm sharing them here)

The result holds in any unital C$^*$-algebra. So assume that $\|p-q\|<1$, with $p,q$ in a unital C$^*$-algebra $A\subset B(H)$.

Claim 1: There is a continuous path of projections joining $p$ and $q$.

Proof. Let $\delta\in(0,1)$ with $\|p-q\|<\delta$. For each $t\in[0,1]$, let $x_t=tp+(1-t)q$. Then $$ \|x_t-p\|=\|(1-t)(p-q)\|<\delta(1-t), $$ $$ \|x_t-q\|=\|t(p-q)\|<\delta t. $$ This, together with the fact that $x_t$ is selfadjoint, implies that $\sigma(x_t)\subset K=[-\delta/2,\delta/2]\cup[1-\delta/2,1+\delta/2]$ (since $\min\{t,1-t\}\leq1/2$). Now let $f$ be the continuous function on $K$ defined as $0$ on $[-\delta/2,\delta/2]$ and $1$ on $[1-\delta/2,1+\delta/2]$. Then, for all $t\in[0,1]$, $f(x_t)\in A$ is a projection. And $$ t\to x_t\to f(x_t) $$ is continuous, completing the proof of the claim. Edit: years later, I posted this answer to a question on MSE that proves the continuity.

Claim 2: We may assume without loss of generality that $\|p-q\|<1/2$.

This is simply a compacity argument, using that each projection in the path $f(x_t)$ is very near another projection in the path. The compacity allows us to make the number of steps finite, and so if we find projectons $p=p_0,p_1,\ldots,p_n=q$ and unitaries with $u_kp_ku_k^*=p_{k+1}$, we can multiply the unitaries to get the unitary that achieves $q=upu^*$.

Claim 3: If $\|p-q\|<1/2$, there exists a unitary $u\in A$ with $q=upu^*$.

Let $x=pq+(1-p)(1-q)$. Then $$ \|x-1\|=\|2pq-p-q\|=\|p(q-p)+(p-q)q\|\leq2\|p-q\|<1, $$ so $x$ is invertible. Let $x=uz$ be the polar decomposition. Then $u$ is a unitary. Note that $$ qx^*x=q(qpq+(1-q)(1-p)(1-q))=qpq, $$ so $q$ commutes with $x^*x$ and then with $z=(x^*x)^{1/2}$. Note also that $px=xq$, so $puz=uzq=uqz$. As $z$ is invertible, $pu=uq$, i.e. $$ q=u^*pu. $$ Note that $u=xz^{-1}\in A$.

Qiaochu give a "real" proof that works in any C*-algebra. If all you care about is $B(H)$, we can give a very simple proof because all you need for $p$ and $q$ to be unitarily equivalent is that their ranges have the same dimension, and their kernels have the same dimension. Well, if two closed subspaces have different dimensions then the one with larger dimension intersects the orthocomplement of the other [proof below], and this easily implies that $\|p - q\| < 1$ forces $p$ and $q$ to have ranges and kernels with matching dimensions.

[Proof of the claim: first suppose $E$ and $F$ are subspaces of a finite dimensional space and ${\rm dim}(E) < {\rm dim}(F)$. Then $n < {\rm dim}(E^\perp) + {\rm dim}(F) = {\rm dim}(E^\perp \vee F) + {\rm dim}(E^\perp \wedge F)$ where $n$ is the dimension of the ambient space. So ${\rm dim}(E^\perp \wedge F) > 0$. Now if $E$ and $F$ are subspaces of a separable infinite dimensional space and ${\rm dim}(E) < {\rm dim}(F)$ then ${\rm dim}(E)$ is finite and we can find a finite dimensional subspace $F_1$ of $F$ with ${\rm dim}(F_1) < {\rm dim}(E)$. Then run the preceding argument within the finite dimensional space $E \vee F_1$.]