Without loss of generality, $u_1,\dots,u_n$ can be taken to be the standard basis of ${\bf R}^n$ (so $e_1,\dots,e_r$ is just some arbitrary orthonormal system). We can view $\Lambda$ as a real symmetric $n \times n$ matrix of Frobenius norm $1$ and rank at most $r$. (Conversely, every such matrix has a representation of the desired form for some $e_1,\dots,e_r$ by the spectral theorem, so this reformulation has not lost any information.) If $D = \sum_{i=1}^n \langle \Lambda, u_i \otimes u_i \rangle u_i \otimes u_i$ is the diagonal component of $\Lambda$, the hypothesis is then $\|D\|^2 \geq 1-\delta$.

Let $\lambda = \mathrm{diag}(\lambda_1,\dots,\lambda_r,0,\dots,0)$ be the diagonalisation of $\Lambda$ (the ordering of the eigenvalues is irrelevant), thus $\lambda$ is a unit vector supported on a set of $r$ indices in $\{1,\dots,n\}$. By the Schur-Horn theorem, $D$ is a convex combination of the permutations $\sigma(\lambda)$ of $\lambda$, $\sigma \in S_n$, where the symmetric group $S_n$ acts on diagonal matrices in the obvious manner, thus
$$ D = \sum_{\sigma \in S_n} c_\sigma \sigma(\lambda)$$
for some non-negative coefficients $c_\sigma$ summing to $1$. (Again, the Schur-Horn theorem is an if-and-only-if statement, so we have still not lost any information so far.)

Now we exploit the uniform convexity of the Frobenius norm. We take the norm square
$$ 1-\delta \leq \|D\|^2 = \sum_{\sigma,\sigma' \in S_n} c_\sigma c_{\sigma'} \langle \sigma(\lambda), \sigma'(\lambda) \rangle$$
and then apply the cosine rule to conclude
$$ \sum_{\sigma,\sigma' \in S_n} c_\sigma c_{\sigma'} \| \sigma(\lambda) - \sigma'(\lambda)\|^2 \leq 2\delta$$
hence by pigeonholing there exists $\sigma_0 \in S_n$ such that
$$ \sum_{\sigma \in S_n} c_\sigma \| \sigma(\lambda) - \sigma_0(\lambda)\|^2 \leq 2\delta$$
hence by Cauchy-Schwarz
$$ \sum_{\sigma \in S_n} c_\sigma \| \sigma(\lambda) - \sigma_0(\lambda)\| \leq \sqrt{2\delta}$$
hence by the triangle inequality
$$ \| \sum_{\sigma \in S_n} c_\sigma \sigma(\lambda) - \sigma_0(\lambda)\| \leq \sqrt{2\delta}$$
thus
$$ \| D - \sigma_0(\lambda) \| \leq \sqrt{2\delta}.$$
The diagonal matrix $\sigma_0(\lambda)$ is supported on a set $T \subset \{1,\dots,n\}$ of cardinality $r$, and the above inequality implies that the Frobenius norm of $D$ outside of $T$ is at most $\sqrt{2\delta}$. Thus
$$ \sum_{i \not \in T} |\langle \Lambda, u_i \otimes u_i \rangle|^2 \leq 2\delta$$
so that
$$ \sum_{i \in T} |\langle \Lambda, u_i \otimes u_i \rangle|^2 \geq 1-3\delta$$
as required.