Countability of the set of $t$ such that $E-tB$ is not injective Consider a separable Hilbert space $\mathcal H$ and two compact self adjoint continuous operators $E,B:\mathcal H \to \mathcal H$. $E$ is injective.
Now consider the set
$$\tau = \{t\in [0,1]: E-tB\;\; \mathrm{is}\;\mathrm{NOT}\; \mathrm{injective} \}\,.$$
I think that the cardinality of $\tau$ is at most countable.
Clearly everything if easy if $E$ and $B$ share a basis of eigenfunctions. Indeed, since for all $t$ $E-tB$ is compact and self-adjoint, we can write
$$E-tB = \sum_{n=0}^\infty (\lambda_n - t\mu_n)P_n$$
where $P_n$ are the projectors on the eigenspaces and $\lambda_n$ and $\mu_n$ the eigenvalues of $E$ and $B$ respectively. So in this case $$\tau = \{t\in [0,1]: \lambda_n = t \mu_n\;\;\mathrm{for}\;\mathrm{some}\;n\}\,,$$
which is clearly countable.
But how to deal with the general case? The property does still hold true?
 A: Here is a partial answer in the special case that $E\geq 0$.
Assume by contradiction that $\{t_i\}_{i\in I}$ is an uncountable subset of $[0,1]$ such that $E-t_iB$ is not injective for
all $i$, and choose nonzero vectors $\{x_i\}_{i\in I}\subseteq H$ such that $E(x_i)=t_iB(x_i)$, for all $i$.
If $t_i\neq t_j$ notice that
$$
  t_i\langle B(x_i),x_j\rangle  =
  \langle E(x_i),x_j\rangle  =
  \langle x_i,E(x_j)\rangle  =
  \langle x_i,t_jB(x_j)\rangle =
  t_j\langle B(x_i),x_j\rangle ,
  $$
so $\langle B(x_i),x_j\rangle  =0$, and consequently also   $\langle E(x_i),x_j\rangle =0$.  Using that $E$ is positive we may write
$E=E^{1/2}E^{1/2}$, so
$$
  0=\langle E(x_i),x_j\rangle  = \langle E^{1/2}(x_i),E^{1/2}(x_j)\rangle ,
  $$
and it follows that
$\{E^{1/2}(x_i)\}_{i\in I}$ is an uncountable family of pairwise orthogonal vectors in $H$, a contradiction.

EDIT (1): Here is another interesting fact.  Should the answer to the original question  be affirmative then it is also
affirmative without the hypothesis that $B$ and $E$ are self-adjoint.
Here is why:  suppose that the (possibly non-selfadjoint) compact operators
$B$ and $E$, with $E$ injective, yield a counter-example, that is,
one can find an uncountable subset
$\{t_i\}_{i\in I}\subseteq [0,1]$ and a corresponding family $\{x_i\}_{i\in I}\subseteq H$ of nonzero vectors such that $E(x_i)=t_iB(x_i)$, for all $i$.
Consider the operators $\tilde B$ and $\tilde E$, acting on $H\oplus H$, defined as follows:
$$
  \tilde B = \pmatrix{0 & B^*\cr B & 0}, \qquad \tilde E = \pmatrix{0 & E^*\cr E & I},
  $$
where $I$ denotes the identity operator on $H$.
Also consider the vectors $\tilde x_i\in H\oplus H$ given by $\tilde x_i = \pmatrix{x_i\cr 0}$.
An easy computation shows that $\tilde E(\tilde x_i)=t_i\tilde B(\tilde x_i)$, so $\tilde E-t_i\tilde B$ is not
injective.  Evidently both $\tilde B$ and $\tilde E$ are compact and self-adjoint, and we will next  show that $\tilde E$ is
injective.  For this suppose that $\pmatrix{x\cr y}$ lies in the null space of $\tilde E$.  It therefore follows that
$E^*(y) = 0$ and $E(x)+y=0$.  Applying $E^*$ to the latter identity gives
$$
  0 = E^*E(x)+E^*y=E^*E(x),
  $$
hence
$$
  0 = \langle E^*E(x), x\rangle  = \langle E(x), E(x)\rangle  = \Vert E(x)\Vert ^2,
  $$
leading to $E(x)=0$, and also  $x=0$, because $E$ is injective.  Plugging this into $E(x)+y=0$, finally gives $y=0$, as
well.
Therefore the pair $(\tilde B, \tilde E)$ provides a counter-example for the original question which we are assuming to
have a positive answer.  We have thus come to a contradiction, hence proving the statement.

EDIT (2): The compactness hipotheses can be removed as well!!   Here is why: suppose that the (possibly
non-compact) bounded operators $B$ and $E$, with $E$ injective, yield a counter-example, that is, one can find an uncountable
subset $\{t_i\}_{i\in I}\subseteq [0,1]$ such that
$E-t_iB$ is not injective for all $i$.
Every separable Hilbert space admits an injective compact operator (e.g. the diagonal operator with diagonal entries
$1,1/2,1/3,\ldots $  on $l^2$) so let $K$ be such an operator on $H$.  Then clearly $KE$ is injective but
$$
  KE-t_iKB = K(E-t_iB)
  $$
is not.
Thus the pair of compact operators  $(KB, KE)$ provides a counter-example as in EDIT (1) which in turn
can be made into a counter-example for the original question.

EDIT (3): In this post one
will find a counter-example for the situation in EDIT (2) so the question is finally settled in the NEGATIVE!!
To be a
bit more specific, $E$ is taken to be the identity operator and $B$ the backward shift (mind you that for $t$ nonzero
one has that $E-tB$ is injective iff $t^{-1}E-B$ is).
