$U\subset X$ infinite-dimensional subspace with $(U,\Vert\cdot\Vert)$ complete. Is $(X,\Vert\cdot\Vert)$ also complete with the same norm? As I came across that completeness is equivalent to closeness of a subspace if the whole space is complete, I was wondering if I can prove or disprove that the completness of a subspace also can give completness of the whole space. 
One problem was, as I tried different approaches, that the norm was not a valid one for the bigger space. I also tried artificial examples as $C([0,1])$ with the $L_2$-norm that is not complete to intersect with some $L_p$ space to get something. My intuition is telling me that it shouldn't be true.
Can someone help me or give me a hint? Thanks in advance!
EDIT: Sorry, I forgot to mention that the subspace should be NOT finite. Because, as Lord Shark the Unknown mentioned, every finite-dimensional subspace of a normed space is complete.
 A: There is indeed a counterexample.
Take the field of $p$-adic numbers $\mathbb Q_p$, which is complete with the usual $p$-adic norm $\vert\vert \cdot\vert\vert_p$ (by construction if you construct it right).
Then the algebraic closure $\mathbb Q_p^a$ of $\mathbb Q_p$ is not complete.
Its completion is $\widehat {\mathbb Q_p^a}$ and it's called Tate's field.
A: The example is from this question.
Consider the Banach space $\ell^2(\mathbb{Z})$ and define its subspace:
$$X = \{(x_n)_{n\in\mathbb{Z}} \in \ell^2(\mathbb{Z}) : x_n \ne 0 \text{ for at most finitely many } n < 0\}$$
$(X, \|\cdot\|_2)$ is not a Banach space as it is not a closed subspace of $\ell^2(\mathbb{Z})$.
However $$U = \{(x_n)_{n\in\mathbb{Z}}  \in \ell^2(\mathbb{Z}): x_n = 0 \text{ for all } n < 0\} \le X$$ 
is a complete infinite-dimensional subspace of $X$, because obviously $U \cong \ell^2(\mathbb{N})$.
A: Let $U$ be  any Banach space with norm $\|\cdot\|_U.$  Let $V$ be a normed linear space that is not complete, with norm $\|\cdot \|_V$  and origin $0_V.$
Let $X=U\times V$ with $\|(u,v)\|_X=\|u\|_U+\|v\|_V.$ Then $U\times \{0_V\}$ is a closed linear subspace of $X,$ and is complete because it is isometrically isomorphic to $U.$ But $X$ is not complete.
