linear subspace of dual space $X$ is a locally convex space and $X^*$ is its dual space with weak* topology or uniform topology. If $H$ is a linear subspace of $X^*$ such that $\bar H  \ne X^*$, then is there a non-zero $x \in X$ such that $f(x)=0$ for all $f \in H$? 
 A: For the weak$^\ast$-topology the answer is yes.
Two very similar ways of seeing it, both using the fact(*) that $\psi\colon X^\ast \to \mathbb{C}$ is weak$^\ast$-continuous if and only if there is $x \in X$ such that for all $f \in X^\ast$ we have  $\psi(f) = f(x)$: $\DeclareMathOperator{\re}{Re}\DeclareMathOperator{\ev}{ev}$


*

*Let $g \in X^\ast \smallsetminus \bar{H}$. By the Hahn-Banach separation theorem there exists a weak$^\ast$-continuous linear functional $\psi: X^\ast \to \mathbb{C}$ such that $\re\psi(h) \lt \re\psi(g)$ for all $h \in \bar{H}$ since $\bar{H}$ is closed and convex and $g \notin \bar{H}$. Since $\bar{H}$ is a linear subspace, it follows that $\psi(h) = 0$ for all $h \in \bar{H}$. But there is $x \in X$ such that $\psi(f) = f(x)$ and $\re{g}(x) = \re\psi(g) \gt 0$, so $x \neq 0$.

*The quotient space $Q = X^\ast / \bar{H}$ is a non-zero Hausdorff locally convex vector space and $\pi\colon X^\ast \to X^\ast/\bar{H}$ is continuous. By Hahn–Banach $Q$ admits a non-zero continuous linear functional $\varphi \in Q^\ast$. Since $\pi$ is onto, $\varphi \circ \pi \in X^\ast$ is non-zero and it is continuous as a composition of continuous maps. By definition $\varphi \circ \pi$ annihilates $\bar{H}$ and since every weak$^\ast$-continuous linear functional on $X^\ast$ is of the form $f \mapsto f(x)$ for some $x \in X$, we're done: $\varphi\circ\pi(f) = f(x)$ for some $x \in X$ and $x$ can't be zero because $\varphi\circ\pi$  is non-zero, and we have $h(x) = (\varphi\circ\pi)(h) = 0$ for all $h \in \bar{H}$.

(*) To see this, note that weak$^\ast$-continuity of $\psi$ implies that there are points $x_{1},\ldots, x_{n} \in X$ such that
$$
\lvert\psi(f)\rvert \leq \max_{i=1,\ldots,n}\lvert f(x_i)\rvert
$$
for all $f \in X^{\ast}$. This shows in particular that $\bigcap_{i = 1}^{n} \ker(\ev_{x_{i}}) \subset \ker \psi$ and this happens if and only if $\psi$ is a linear combination of the $\ev_{x_i}$: consider the subspace
$$
V =
\left\{
  \big(f(x_1), \ldots, f(x_n)\big)
  \,:\, f \in X^{\ast}
\right\} \subset \mathbb{C}^{n}. 
$$
The functional $\tilde{\psi}: V \to \mathbb{C}$ given by
$\left(f(x_1), \ldots, f(x_n)\right) \mapsto \psi(f)$ is well-defined and thus it extends to a linear functional of the form
$(v_1,\ldots,v_{n}) \mapsto \sum_{i=1}^{n} \alpha_{i} v_{i}$, which shows that $\psi =  \sum_{i=1}^{n} \alpha_{i} \ev_{x_i} = \ev_{\alpha_{1}x_{1} + \cdots + \alpha_{n}x_{n}}$, so $\psi = \ev_x$ with $x = \alpha_{1}x_{1} + \cdots + \alpha_{n}x_{n}$.

Added:
Now that the terminology on the uniform topology was clarified, let me answer this part negatively.
Let $X^\ast$ be the dual space of a non-reflexive Banach space $X$, e.g. $X = c_0$. The uniform topology on $X^\ast$ is simply the norm-topology on $X^\ast$ because every seminorm $p(f) = \sup_{x \in B}|f(x)|$ with bounded $B \subset X$ is dominated by (or equal to) a multiple of the norm $\lVert\,\cdot\,\rVert_{X^\ast}$. Thus, every element $\varphi \in X^{\ast\ast} \smallsetminus X$ will be continuous with respect to the uniform topology, but it won't be weak$^\ast$-continuous by (*) because $\varphi \neq \ev_x$ for all $x \in X$. Since a linear functional is continuous if and only if its kernel is closed, $H = \ker{\varphi}$ is uniformly closed, but not weak$^\ast$-closed. It follows that the weak$^\ast$-closure of $H$ is all of $X^\ast$, and the only linear functional of the form $\ev_x$ vanishing on all of $H$ is the zero functional.
