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Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$, with $\|v_i\|=1$ for all $i$. To each $i$ associate a positive real number $\lambda_i>0$ such that $\forall\varepsilon>0$ there is $i\in I$ such that $\lambda_i<\varepsilon$.

  Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-\lambda_i,\lambda_i]\forall i\in I\right\}$. It has empty interior because for all $v\in X$ and all $i$, $v+3\lambda_iv_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$ with $\inf_{i\in I}\|v_i\|=0$. Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-1,1]\forall i\in I\right\}$. It has empty interior because for all $v\in X$ and all $i$, $v+3v_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not. This convex set $X$ is necessarily not closed, as implied by user1286767's answer.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$, with $\|v_i\|=1$ for all $i$. To each $i$ associate a positive real number $\lambda_i>0$ such that $\forall\varepsilon>0$ there is $i\in I$ such that $\lambda_i<\varepsilon$.

Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-\lambda_i,\lambda_i]\forall i\in I\right\}$. It has empty interior because for all $v\in X$ and all $i$, $v+3\lambda_iv_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$, with $\|v_i\|=1$ for all $i$. To each $i$ associate a positive real number $\lambda_i>0$ such that $\forall\varepsilon>0$ there is $i\in I$ such that $\lambda_i<\varepsilon$.

Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-\lambda_i,\lambda_i]\forall i\in I\right\}$. It has empty interior because for all $v\in X$ and all $i$, $v+3\lambda_iv_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.

Here is an example of a convex subset $X$ of an infinite dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not. This convex set $X$ is necessarily not closed, as implied by user1286767's answer.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$, with $\|v_i\|=1$ for all $i$. To each $i$ associate a positive real number $\lambda_i>0$ such that $\forall\varepsilon>0$ there is $i\in I$ such that $\lambda_i<\varepsilon$.

Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-\lambda_i,\lambda_i]\forall i\in I\right\}$. It has nonempty interior because for all $v\in X$ and all $i$, $v+3\lambda_iv_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not. This convex set $X$ is necessarily not closed, as implied by user1286767's answer.

Let $(v_i)_{i\in I}$ be a Hamel basis of $H$, with $\|v_i\|=1$ for all $i$. To each $i$ associate a positive real number $\lambda_i>0$ such that $\forall\varepsilon>0$ there is $i\in I$ such that $\lambda_i<\varepsilon$.

Our convex subset is $X:=\left\{\sum_{i\in I}a_iv_i;a_i\in[-\lambda_i,\lambda_i]\forall i\in I\right\}$. It has empty interior because for all $v\in X$ and all $i$, $v+3\lambda_iv_i$ is not in $X$. And it is not contained in any hyperplane because for any $v\in H$, $\varepsilon v\in X$ for small enough $\varepsilon>0$.