If every polynomial in $k[x]$ has a root in $E$, is $E$ algebraically closed? If $E/k$ is algebraic and for all $f$ in $k[X]$, all roots of $f$ lie in $E$, then $E$ is algebraically closed.
The question is:
If $E/k$ is algebraic and for all $f$ in $k[X]$, at least one root of $f$ lies in $E$, then is $E$ algebraically closed?

This is true, but it is not trivial. See Gilmer, A Note on the Algebraic Closure of a Field. 

The OP asked for another reference in the comments. A google search reveals Richman A theorem of Gilmer and the canonical universal splitting ring, which apparently gives a constructive proof. In this Math Stackexchange answer, Martin Brandenburg gives as an additional reference Isaacs Roots of Polynomials in Algebraic Extensions of Fields. 
Isaacs proves a generalization of Gilmer's theorem: An algebraic extension $K$ of a field $k$ is determined up to isomorphism over $k$ by the set of polynomials in $k[x]$ which have a root in $K$. He cites Gilmer and p.88 of a book called Theory of Fields by Nagata. It's not clear to me that such a book exists, but I did track down a proof of Gilmer's theorem as Theorem 2.12.2 on p. 71 of Nagata's Theory of Commutative Fields. Regarding Gilmer's theorem, Isaacs writes: 

This theorem is not quite the triviality it may appear to be at first glance. If one knows that all polynomials in $F[X]$ split over $E$, then it is an easy exercise to show that $E$ is algebraically closed. Under the weaker hypothesis of Theorem 1, however, this conclusion is considerably more difficult to prove. (It is more difficult to find in the literature, too. A search of about a dozen books that deal with field extensions was able to uncover only one proof of this result and two cases where at least a part of Theorem 1 appears as a  problem.)

