Definition
Let be $X$ a topological vector space. A subset $S$ of $X$ is said convex if the affine combination $$ A:=\{z\in X: z=(1-t)x+ty, t\in[0,1]\} $$ is contained in $S$ for any $x, y\in S$.
Statement
If $S$ is convex and $\text{int}(S)\neq\emptyset$ then $\text{cl}\big(\text{int}(S)\big)=\text{cl}(S)$.
Clearly $\text{int}(S)\subseteq S$ and so $\text{cl}\big(\text{int}(S)\big)\subseteq\text{cl}(S)$. Then if $x\in\text{cl}(S)$ then or $x\in\text{int}(S)$ and so $x\in\text{cl}\big(\text{int}(S)\big)$ either $x\in\text{Bd}(S)$ and so we only to prove that $\text{Bd}(S)\subseteq\text{cl}\big(\text{int}(S)\big)$ but unfortunately I don't be able to prove this. So could someone help me, please?
The answer to Topological properties of convex sets shows that if $K_1$ is closed and convex and $\operatorname{int}(K_1) \ne \emptyset$, then $\operatorname{cl}(\operatorname{int}(K_1)) = K_1$. The present question asks to prove the following more general theorem:
$\operatorname{cl}(\operatorname{int}(S)) \subset \operatorname{cl}(S)$ is obvious. It remains to prove that if $b \in \operatorname{cl}(S)$, then $b \in \operatorname{cl}(\operatorname{int}(S))$. This means to show that for each open neighborhood $U$ of $b$ in $X$ there exists $b_U \in U \cap \operatorname{int}(S)$.
We know that there exists $c_U \in U \cap S$. Fix $a \in \operatorname{int}(S)$. The set $(1-t)\operatorname{int}(S)$ is open in $X$ for $t \ne 1$ because multiplication by a non-zero factor is a homeomorphism on $X$. Hence $\Omega(t) = (1-t)\operatorname{int}(S) + tc_U$ is open in $X$ for $t \ne 1$ because each translation is a homeomorphism on $X$. We have
$(1-t)a + t c_U \in \Omega(t)$.
$\Omega(t) \subset S$ for $t \in [0,1]$ since each point of $\Omega(t)$ has the form $(1-t)c + tc_U$ with $c \in \operatorname{int}(S) \subset S$ and $c_U \in S$.
$\Omega(t) \subset \operatorname{int}(S)$ for $t \in [0,1)$ beause $\Omega(t)$ is an open subset of $S$ for $t \ne 1$.
The function $\gamma : [0,1] \to X, \gamma(t) = a + t(c_U-a) = (1-t)a + tc_U$, is continuous with $\gamma(1) = c_U \in U$, thus $\gamma(t) \in U$ for $t \in (1-\epsilon,1]$. Hence for $t \in (1-\epsilon,1)$ we see that $b_U = (1-t)a + tc_U \in U \cap \Omega(t) \subset U \cap \operatorname{int}(S)$. Note that the constant maps $\phi(t) = a$ and $\psi(t) = c_U-a$ are continuous and that $\iota(t) = 1-t$ is continuous. Thus 3. and 4. below show that $\gamma$ is continuous.
Edited:
A (real or complex) topological vector space is vector space $X$ endowed with a topology such that addition $\alpha : X \times X \to X, \alpha(x,y ) = x+y$, and scalar multiplication $\mu : \mathbb K \times X \to X, \mu(r,x) = r \cdot x$, are continuous. Here $\mathbb K = \mathbb R, \mathbb C$. It is an easy exercise to show that normed linear spaces are topological vector spaces (the topology is of course the norm-topology).
Multiplication by a non-zero factor is a homeomorphism on $X$: The map $i_r : X \to \mathbb K \times X, i_r(x) = (x,r)$, is continuous for all $r \in \mathbb K$. Hence $\mu_r = \mu \circ i_r : X \to X$ is continuous. If $r \ne 0$, then also $\mu_{1/r} : X \to X$ is continuous. Clearly $\mu_r \circ \mu_{1/r} = id$ and $\mu_{1/r} \circ \mu_r = id$, thus $\mu_r$ is a homeomorphism with inverse $\mu_{1/r}$.
Each translation is a homeomorphism on $X$: The map $j_b : X \to X \times X, j_b(b) = (x,b)$, is continuous for all $b \in X$. Hence $\alpha_b = \alpha \circ j_b : X \to X$ is continuous. Clearly $\alpha_b \circ \alpha_{-b} = id$ and $\alpha_{-b} \circ \alpha_b = id$, thus $\alpha_b$ is a homeomorphism with inverse $\alpha_{-b}$.
If $r : Y \to \mathbb K$ and $f : Y \to X$ are continuous, then $r\cdot f : Y \to X, (r \cdot f)(y) = r(y) \cdot f(y)$, is continuous: The diagonal map $d : Y \to Y \times Y, d(y) = (y,y)$, is continuous. Thus $r \cdot f = \mu \circ (r \times f) \circ d$ is continuous.
The sum of continuous functions $f,g: Y \to X$ is continuous: We have $f + g = \alpha \circ (f \times g) \circ d$.